User alireza olfati - MathOverflow

Maximal ideals of the rings of Baire- One Functions

2012-08-06T20:11:25Z

A real function $f:X\rightarrow \mathbb{R}$ Is called Baire-one function, if there is a sequence $(f_n)_{n=1} ^\infty$ of continuous functions $f_n:X\rightarrow \mathbb{R}$ on $X$ so that for all $x\in X$ $$lim_{n\rightarrow \infty}f_n(x)=f(x)$$

When $X$ is a banach space, we have the following theorem refered to as Baire factorization theorem.

Theorem:The real function $f:X\rightarrow \mathbb{R}$ is in the class of baire-one if and only if for all closed subset $K\subset X$, the restricted function $f|_K$ has a point of continuity with respect to $K$.

Definition: We denote the set of all baire-one real functions on the space $X$ by $Ba_1(X)$.

As you could easily see , $Ba_1(X)$ forms a ring with pointwise addition and multiplication. for simplicity Let me consider $X=[0 , 1]$.

suppose $C[0 , 1]$ denotes the ring of all continuous real valued functions on the interval $[0 , 1]$. by the theorem of Gelfond and Kolmogroff we Know that the set of all maximal ideals of the ring $C[0 , 1]$ is of the form {$M_x: x\in X$} ,Where $M_x=${$f\in C[0, 1]: f(x)=0$}.

Compared with the ring $C[0 , 1]$ we could easily find that the sets of the form $M_x^1=$

{$f\in Ba_1[0 , 1]: f(x)=0$} are maximal ideals of the ring $Ba_1[0 , 1]$. From this property some Questions came in my mind as follows:

Question1: Does there exist a maximal ideal in $Ba_1[0 , 1]$ other than maximal ideals of the form $(M_x^1$ for $x\in X)$

Question2: Is the ring $Ba_1[0 , 1]$ a PM- ring?$($i.e. a ring in which each prime ideal is contained in a unique maximal ideal.$)$

Compact subsets and Hausdorffness of Topology

2012-06-01T20:03:34Z

We Know that all closed subsets of a compact topological space $(X,\tau)$ are compact. But if we add The Hausdorff condition on the topology $\tau$ we could see the equivalence of these subsets.(i.e. in compact Hausdourff spaces closed subsets are the same as compact subsets)

Know for asking the converse of the above fact we could or not omit the compactness of the space$(X,\tau)$ as follows:

(STATEMENT) If all compact subsets of a topological space $(X,\tau)$ are closed then $(X,\tau)$ is Hausdorff.
If the above statement is not valid, Is there a separation axiom weaker than Hausdorffness on the space $X$ that compact subsets are closed?

For the first statement If we add the condition of compactness of $(X,\tau)$, it changes as follows:

Is The space $(X,\tau)$ Hausdorff,If closed subsets and compact subsets are equivalent in $X$?

Quotient rings of $C(X)$

2012-08-15T15:17:59Z

Let $X$ be a Tychonoff topological space. Consider the ring $C(X)$ of all continuous real valued functions on $X$. For what conditions on an ideal $I$ of $C(X)$, we could deduce that the quotient ring is isomorphic to a ring of the form $C(Y)$.i.e.

Question : An ideal $I$ has the algebraic property $\mathcal{P}$ if and only if The exists a Tychonoff topological space $Y$ so that: $$\frac{C(X)}{I} \cong C(Y)$$

special extremally disconnected spaces with only finite isolated points

2012-11-22T11:39:06Z

We Know that a cardinal $\kappa$ is measurable if there is a set $X$ with cardinal $\kappa$ and a {0,1}-measure $\mu: P(X) \rightarrow ${$0,1$} so that for all $x \in X$, $\mu(x)=0$ and $\mu(X)=1$. also a cardinal which is not measurable, is called non measurable.

Also we Know this is unprovable to find a set with measurable cardinal in "ZFC".

In topology an extremally disconnected space is a topological space in which all open subsets has open closure.

Also we call a topological space to be a P-space if all it's $G_{\delta}$- sets are open.

There is a well-Known theorem that says every extremally disconnected P-space with non measurable cardinal is discrete.

From the aforesaid summaries a question could be posed that:

Question: If we suppose that a measurable cardinal exists, can we construct an extremally disconnected P-space with only a finite number of isolated points.

Answer by AliReza Olfati for Relative extremely disconnected space

2012-11-24T16:49:05Z

Hello dear Ali. I think the answer is yes. consider the closed unit interval $I=[0,1]$, and define the set $K$ as follows:$$K=I\times I -(0,1)\times (0)$$

roughly speaking eliminate the interval $(0,1)$ from the bottom of the unit square.

Now we are to define the base of each point of $K$.

If $(x,y)\neq (0,0) , (1,0)$ define the neighborhoods to be as in the usual Euclidean topology.

If $(x,y)=(0,0)$ define the base to be all the sets $[0,\frac{1}{2})\times (0,\epsilon)$, where $\epsilon>0$.

If $(x,y)=(1,0)$ define the base to be all the sets $(\frac{1}{2},1]\times(0,\delta)$, where $\delta>0$.

It is obvious to see that this new space is Hausdorff. But it is not relatively extremely disconnected. To see this consider any neighborhoods of $(0,0)$ and $(1,0)$. it is intuitive to see that the closure of these neighborhoods intersect each other in some point at the edge $x=\frac{1}{2}$.

functional subrings

2012-08-28T14:51:37Z

I should recall the notion of maximal subring of a commutative unitary ring $R$.

Def: A commutative ring $S$ is called a maximal subring of $R$ if $S \subset R$ and if $T \subset R$ constitute a commutative ring with the restricted addition and multiplication of $R$ and also $S\subsetneq T$ then we could deduce that $T=R$.

I am interested in studing this notion in Rings of continuous functions.

We could easily deduce that for $x \neq y \in X$ The set of the form $$M_{x,y}=\Big(f\in C(X): f(x)=f(y) \Big)$$ forms a maximal subring of $C(X)$

From the above summary and notations I could pose my Questions.

Question1: Is there a maximal subring in $C(X)$ other than all $M_{x,y}$'s?

Question2: Is $X$ compact if all maximal subrings of $C(X)$ is of the form $M_{x,y}$?

PS:I suppose that all subrings of a commutative ring $R$ contains the unitary element of $R$.

Answer by AliReza Olfati for a questions about the sums of intersections of maximal ideals

2012-10-18T17:31:26Z

Let $\mathcal{M}_f$ be the intersection of all maximal ideals in $C(X)$ which contains $f$. it is easy to show that $$\mathcal{M} _f=\Big(g \in C(X): Z(f) \subseteq Z(g)\Big)$$

Now Let $I$ be a $z-$ideal of $C(X)$. It means if $f \in I$ and $Z(f)\subseteq Z(g)$, then $g \in I$.

It suffices to show that for all $f \in I$, $\mathcal{M}_f \subseteq I$.

let $g \in \mathcal{M}_f$, by the above characterization of $\mathcal{M}_f$ we have $Z(f) \subseteq Z(g)$ and this implies that $g \in I$.

So we have shown that for all $f \in I$, $\mathcal{M} _f \subseteq I$. this yields: $$I=\sum _{f \in I} \mathcal{M} _f$$. And this is the case which you wanted.

At the end let me show the sketch of the characterization of $\mathcal{M} _f$ .

Suppose $g$ is in all maximal ideals which contains $f$. Fix $x \in Z(f)$, then $f \in M_x$. but $M_x$ is a maximal ideal which contains $f$ .so $g \in M_x$ and this implies that $x \in Z(g)$. then we show that $Z(f) \subseteq Z(g)$.

For the converse let $Z(f) \subseteq Z(g)$. suppose on the cotrary that there exists a maximal ideal $M$ such that $f \in M$ but $g \notin M$.

So we have $M+(g)=C(X)$. this shows that there exists $k \in M$ and $h \in C(X)$ so that $$k+h.g=1$$

This implies that $Z(k) \cap Z(g)= \varnothing$. But we Know that $f, k \in M$. this obviously shows that $Z(f) \cap Z(g) \neq \varnothing$. We came into a contradiction, because we supposed that $Z(f) \subseteq Z(g)$.

Arbitrary small positive lower semi continuous functions

2012-10-05T17:18:47Z

This question is a generalization of the question posed in this page to lower semi continuous functions. so let me describe the Question in the following way.

Def: Let $(X,\tau)$ be a Tychonoff Topological space. we say that this space has an arbitrary small lower semi continuous function, if the following statement is true for it:

Statement:For each $x\in X$ consider an arbitrary positive real number $\epsilon_x>0$. then there exists a lower semi continuous real valued function $f:X\rightarrow \mathbb{R}$ with the following property: $$\forall x \in X $$ $$ 0< f(x) < \epsilon_x$$

Question:Can we characterize the spaces which has the above statement as it's property?

Kaplansky's theorem and Axiom of choice

2012-09-22T06:55:35Z

Kaplansky in his paper titled by Projective Modules gave an important and essential theorem as follow:

Theorem: Let $R$ be a ring, $M$ an $R$-module which is a direct sum of (any number of) countably generated $R$-modules. Then any direct summand of $M$ is likewise a direct sum of countably generated $R$-modules.

But if you could take a look to the pattern of his proof, he applied the well ordering Theorem for proving it.

I am thinking about the relation of his proof with the well ordering theorem. More precisely I am thinking about the answer of the following question:

Question: Is the Kaplansky theorem equivalent with the Axiom of Choice or it can be proved with the weeker axiom(i.e.boolean prime ideal theorem)?

Existence of algebraic closure and Axiom of choice

2012-09-18T16:00:04Z

Possible Duplicates:
Is the statement that every field has an algebraic closure known to be equivalent to the ultrafilter lemma?
algebraic closure of commuting pairs of matrices

we need zorn's lemma for proving that every field $F$ has a unique algebraic closure. but I haven't seen a converse for this important Theorem.

From the above illustration my question is:

Is it true that the existence of The unique algebraic closure is equevalent to axiom of choice$(AC)$?

Sums of Strongly z-ideals

2012-08-16T06:12:45Z

In the rings of continuous functions,i.e.$(C(X))$ an ideal $I$ is called strongly $z$-ideal if it is an intersection of some maximal ideals of $C(X)$. i.e. $$I=\cap_{\alpha \in A} \mathcal{M_{\alpha}}$$ Where each $\mathcal{M_{\alpha}}$ is a maximal ideal of $C(X)$.

Question: If $I$ and $J$ are two strongly $z$-ideals of $C(X)$, Is the ideal $I+J$ strongly $z$-ideal or all of $C(X)$?

PS: If $X$ is a $P$- space, then each ideal of $C(X)$ is strongly $z$- ideal, And conversly if each ideal of $C(X)$ is strongly $z$- ideal than $X$ is a $P$-space.

separability of commutative rings

2012-09-14T14:37:17Z

Before discussing on the main Question I should recall two notions in the area of commutative rings.

By $Max(R)$, we mean the set of all maximal ideals of the commutative unitary ring $R$.

Definition 1: A commutative ring $R$ is called separable, if for any subset $(M_{\alpha})_{\alpha \in A} \subset Max(R)$ with $\cap _{\alpha \in A}M _{\alpha}=(0)$, there exists a countable subset $A_0 \subset A$ that $\cap _{\alpha \in A _{0}}M _{\alpha}=(0)$.

and

Definition 2: A commutative ring $R$ is called $\aleph_{1}$-cogenerated if for any collection $(I_{\alpha})_{\alpha \in A}$, with $\cap _{\alpha \in A} I _{\alpha}=(0)$ there exists a countable subset $A_0 \subset A$ so that $\cap _{\alpha \in A _{0}}I _{\alpha}=(0)$.

It is clear that every $\aleph_1$-cogenerated ring is separable. I am looking for a commutative $J$-semisimple ring $R$ $($i.e. $J(R)=0)$ that is a separable ring but is not $\aleph_1$-cogenerated.

Question: Is there a commutative $J$-semisimple ring $R$ that is separable but is not $\aleph_1$-cogenerated.

Addendum: The ring $C(X)$ of all continuous real valued functions on a topological space $X$ is an example of a commutative $J$-semisimple ring for which separability and $\aleph_1$-cogenerated are equivalent.

Non trivial definition of bicontinuous functions and the ring of all bicontinuous functions.

2012-08-03T19:29:09Z

At first let me recall that if There are two topology $\tau_1$and $\tau_2$ on a set $X$, the triple $(X,\tau_1,\tau_2)$ is called a bitopological space.

There are many definitions and properties which have been proved for bitopological spaces. The reason which I wrote this note for, was the difficulties of defining a special continuity on bitopological spaces. As you Know there are a lot of definitions for defining bicontinuous functions on bitopological spaces. But there is no suitable definition for bicontinuous functions which the collection $C(X,\tau_1,\tau_2)$ of all bicontinuous real funcutions on bitopological space $(X,\tau_1,\tau_2)$ into real numbers $\mathbb{R}$, becomes a ring.

I thought about the following definition:

...................................................................................................................................................................

Def: $f\in C(X,\tau_1,\tau_2) $ is bicontinuous at $x\in X$, if for all $\epsilon>0$ there are $U\in \tau_1$ and $V\in \tau_2$ so that $$x\in U\cap V, f(U\cap V)\subset (f(x)-\epsilon, f(x)+\epsilon)$$ and obviously $f$ is bicontinuous on $X$, if $f$ is bicontinuous at all $x\in X$.

...................................................................................................................................................................

In this definition $ C(X,\tau_1,\tau_2) $ is a ring but with a closer look at this, we have found nothing new, because with this definition $C(X,\tau_1,\tau_2)$ is exactly the ring $C(X,\tau_1 \vee \tau_2 )$ of all continuous real valued functions on topological space $(X,\tau_1 \vee \tau_2 )$. Now here is my question:

Question: Is there a nontrivial definition of bicontinuous real valued functions so that the collection $C(X, \tau_1, \tau_2)$ would be a ring which is not in general isomorphic to $C(Y,\tau)$ of all continuous real valued functions on some topological space$(Y,\tau)$?

Answer by AliReza Olfati for A space in which sequences have unique limits but compact sets need not be closed

2012-09-07T11:28:10Z

I refer to COROLLARY 1 of This Article.

In COROLLARY 1 of it, $X^+$ denotes the one point compactification of the topological space $X$:

COROLLARY: Let $X$ be a Hausdorff space.Then:

(a) $X^+$ is always $US$.

(b) $X^+$ is $KC$ if and only if $X$ is a $\kappa$ space.

So it suffices to choose a Hausdorff space $X$, which is not $\kappa$ space. then $X^+$ is $US$ but is not $KC$.

PS: The topological space $X$ is called $\kappa$ space, if A subspace $A$ is closed in $X$ if and only if $A \cap K$ is closed in $K$, for all compact subset $K \subset X$.

The set of Upper semi-continuous functions as a ring.

2012-09-02T17:24:35Z

I should recall that the surgenfery topology on the real numbers is denoted by $\mathbb{R}_l$, and has the set
{$[a , b): a,b \in \mathbb{R} $} as it's base.

If $X$ is a topological space, an upper semi-continuous real function on $X$ can be interpret as a continuous function from $X$ into $\mathbb{R}_l$.

The set of all upper semi-continuous real functions on $X$ is denoted by $USC(X)$.

We could easily see that if we consider $X=\mathbb{R}$ then the set $USC(\mathbb{R})$ is not a ring with pointwise addition and multiplication. because there is $f \in USC(\mathbb{R}) $ so that $-f \notin USC(\mathbb{R}) $. Indeed it is not a Group.

Question.For what condition(s) on $X$, the set $USC(X)$ constructs a ring structure we the pointwise addition and multiplication?

PS:I am looking for the topological condition(s) $P$ on $X$ so that, $USC(X)$ is a ring iff $X$ has the property $P$.

Thank you so much for noticing to my Question.

Answer by AliReza Olfati for Is every zero-dimensional space with no infinite clopen partition pseudocompact?

2012-07-31T22:23:12Z

I think the answer is yes. to see the sketch of proof you could suppose on the contrary that, $f$ is not bounded and choose $$x_1, x_2,...,x_m,... \in X$$ and$$r_1,r_2,...r_m,...\in \mathbb{R}^+$$ so that $$r_1< |f(x_1)|< r_2<|f(x_2)|< ... r_m< |f(x_m)|< ...$$ Because $X$ is zero-dimensional and $f$ is continuous, for each $m\in \mathbb{N}$ you could find a clopen subset $V_m \subseteq${$x\in X: r_m< |f(x)|< r_{m+1}$}. You could see that for $i\neq j, V_i\cap V_j=\emptyset$. Moreover $$V_0=X-\cup_{i=1}^{\infty}V_{i} $$ which is equal to $\cup_{i=1}^{\infty}(${$x\in X: |f(x)|< r_{i+1}$}$-V_1\cup V_2 \cup... \cup V_i$$)$, and obviously it is open. It yields that $X$ has an infinite partition and this implies a contradiction. because $X$ is $\omega-$ pseudocompact and has no countable partition of clopen sets.

Answer by AliReza Olfati for Continuity of the maxima

2012-07-07T22:49:17Z

In general the answer is no. For showing this claim I must bring two theorems. You could find these theorems in the page 238 of the text "General topology" written by Ryszard Engelking.

The following theorem is due to Isiwata, Nobel, Hager and comfort:

Theorem1: For the Tychonoff spaces $X , Y$ the following are equivalent:

The Projection $p:X\times Y \rightarrow X$ maps zero-sets of $X\times Y$ to closed sets of $X$.
Every bounded continuous function $f:X\times Y \rightarrow \mathbb{R}$ can be continuously extended over $X \times \beta Y$.
For every bounded continuous function $f:X\times Y \rightarrow \mathbb{R}$ the formula $F(x)=sup_{y\in Y}f(x,y)$ defines a continuous function $F:X\rightarrow \mathbb{R}$.

The following theorem is due to "Tamano" which is essential for our claim:

Theorem2: The cartesian product $X\times Y$ of Tychonoff spaces $X , Y$ is pseudocompact if and only if $X$ and $Y$ are pseudocompact and The Projection $p:X\times Y \rightarrow X$ maps zero-sets of $X\times Y$ to closed sets of $X$.

But for showing our claim as you probably Know there is a countably compact space $X$ which the product space $X\times X$ is not even Pseudocompact.(You could find such example in the chapter9 of the text Rings of continuous functions written by Gillman and jerison)

Then because $X$ is countably compact it is also pseudocompact. and because $X\times X$ is not pseudocompact (as we mentioned above) each of the statements in Theorem1 fails. Then for example there is a bounded continuous function $f:X\times X \rightarrow \mathbb{R}$ so that the function $F(x)=sup_{y\in Y}f(x,y)$ is not continuous.

Ring isomorphism $\Phi \colon C(X) \to C(Y)$ and zero dimensionality of $X$

2012-06-15T08:39:45Z

We denote the ring of all continuous real-valued functions on $X$ by $C(X)$. The ring of all bounded continuous real valued functions on $X$ is denoted by $C_b(X)$. One of the goals of the study of $C(X)$ is to connect algebraic properties of the ring $C(X)$ with the topological properties of the space $X$.

For a simple and well-known connection there is the following theorem:

Theorem: The topological space $X$ is connected iff the ring $C(X)$ has no idempotent element other than $0$ and $1$.

My question comes from the relation between ring homomorphisms of the rings $C(X)$, $C(Y)$ and topological spaces $X$, $Y$.

Q1: Let $X$ and $Y$ be two $T_{3\frac{1}{2}}$topological spaces. If $X$ is a zero dimensional space and $C(X)$ is ring isomorphic to $C(Y)$ (i.e. $C(X)\cong C(Y)$) can we deduce that $Y$ is also a zero dimensional space?

(we recall that a topological space $X$ is zero dimensional if it has a base of clopen subsets.)

The same question can be asked by exchanging the ring $C(X)$ with the Banach algebra $C_b(X)$ as follows:

Q2: Let $X$ and $Y$ be topological spaces with the previous assumption. If $X$ is a zero dimensional space and $C_b(X)$ is ring isomorphic to $C_b(Y)$ (i.e. $C_b(X)\cong C_b(Y)$) can we deduce that Y is also a zero dimensional space?

Answer by AliReza Olfati for The ring of continuous real-valued functions on a Stone space

2012-06-23T17:01:55Z

At first Let me recall the notion of a clean ring:

Definition:A commutative ring $R$ is called clean if every element of $R$ is a sum of a unit and an idempotent.

The following theorem is due to F.Azarpanah who first studied The notion of cleanness in rings of continuous real valued functions. you can find the details of it in This Article

Theorem1:The following statements are equivalent:

$C(X)$ is a clean ring.
$C^*(X)$ is a clean ring.
$X$ is strongly zero-dimensional.(i.e. $\beta X$ is zero-dimensional or stone space. )
Every zero-divisor in $C(X)$ is clean.
$C(X)$ has a clean prime ideal.

I think the illustrious part the above theorem is the relation between cleanness of $C(X)$ and Zero-dimensionality of $\beta X$.

Now Lets turn to your Question. I think the following theorem relates stone spaces to the cleanness property of $C(X)$.

Theorem2:Let $X$ be a compact space. then $X$ is a Stone Space if and only if the ring $C(X)$ is a clean ring.

The proof is clear. because in this case $C(X)=C^*(X)$ is clean iff $\beta X=X$ is Zero dimensional or iff $X$ is a Stone Space.

Existence of an arbitrary Small positive continuous real Valued Function

2012-06-19T18:45:57Z

Let $(X,\tau)$ be a Tychonoff Topological space.

For each $x\in X$ consider an arbitrary positive real number $\epsilon_x>0$. Is There a continuous real valued function $f:X\rightarrow \mathbb{R}$ with the following property:

$$\forall x \in X $$

$$0< f(x) < \epsilon_x$$

From the following comment of Edgar, I have Known that the following Question is the main Purpose of posing this Question, which I didn't notice to write it.

Q.For which properties on $(X,\tau)$, we have the above Property? (one of the properties for which the above condition is true is that $(X, \tau)$ be discrete)

Statement: Is the only property "discreteness" of $X$ ?

On One point Lindeloffication of topological spaces

2012-06-17T17:58:06Z

As you Know when we define a topological space to be the one point compactification of the topological space $X$, we look for a compact space $Y$ such that $X\subset Y$ and $X$ is dense in $Y$ and $|Y-X|=1$.

But if we add an Extra condition that The space $Y$ to be compact and Hausdorff, We Must eliminate a lot of spaces and Spacial topological spaces Could be in this family that satisfies the locally compacness property.

Theorem: A topological space $X$ has a Hausdorff one point compactification iff $X$ is locally compact.

But Compared with the above description, I didn't see Any think about the existence of a space with Lindeloff Property except some situation that I Shall describe it as follows:

At first let me define the lindeloffication of a topological space:

Definition: A Lindeloff space $Y$ is a one point lindeloffication of $X$, if $X$ is a dense subspace of $Y$ and $|Y-X|=1$

We Know that any Discrete space $X$ has a Hausdorff One point lindelofficatin which defined as follows:

$\rightarrow$ Add a point $p$ to $X$ and consider the set $Y=X\cup${$p$}. then all points of $X$ are open and the Neighborhoods of $p$ is of the form: $U\cup${$p$} where $|X-U|\leq\aleph_0$.

Now my Questions are as follows:

$Q_1$: For What condition on the space $X$, It has a Hausdorff One point lindeloffication?

$Q_2$: Is there an obvious example of space $X$ which is non discrete and has a one point lindeloffication?

Added Note: When I posed this Question, I didn't notice that It can be occur for a Hausdorff space to Have More than so-called "One point lindeloffication". Gerald Edgar and David Feldman warned to me that this notion is not Functorial.

Its very important to notice that Compactness implies that having one point compactification is functorial or unique up to Homeomorphism. Let me recall the following theorem:

Theorem1:For $X$ the following Are equivalent:

$X$ is maximal compact.
Every compact subset of $X$ is closed.
Any continuous bijection $f$ from a compact space $Y$ onto $X$ is a homeomorphism.

For lindelof condition we have the same theorem:

Theorem2:For $X$ the following Are equivalent:

$X$ is maximal Lindelof.
The set of all closed subsets of $X$ coincides with the set of all Lindelof subspaces of$X$.
If $Y$ is a lindelof space and $f$ is a continuous bijection from $Y$ onto $X$, Then $f$ is a Homeomorphism.

For the sake of theorem 2 We can find that the one point compactification of an uncountable set is not maximal lindelof.

But We could fix the uniqueness in Question with the maximal lindelofness property as follows:

$Q_3$: For which topological space, we have a maximal Hausdorff one point lindelofication.

Answer by AliReza Olfati for When is a topological group Hausdorff (separated)?

2012-06-13T20:02:59Z

You could find a routin proof in the book "Topological Ring" written by Seth Warner. In this book at page 21 in Theorem 3.4 you could see the following Proposition:

Theorem: Let $G$ be a topological group. The following statements are equivalent:

{$0$} is closed.
{$0$} is an intersection of the neighborhoods of zero.
$G$ is Hausdorff.
$G$ is regular.

You could also find the improvement of it in the book "Topology for analysis" Written by Albert Wilansky. In section 12 at page 243 You could see the following theorem:

THEQREM: Every topological group is completely regular. The following conditions on a topological group $G$ are equivalent:

$G$ is a $T_0$ space.
$G$ is a Tychonoff space.-
$\cap${$U:U$ is a is a neighborhood of $e$}={$e$}

The Proof of Complete regularity Has more details. I think The proof is in the level of Urysohn Lemma.

But if you are interested in the general case, I Suggest You look at the section "uniformity" which were discussed in the following books:

Topology for Analysis: Chapter 11
General topology: Stephen Willard: chapter 9

On Pseudo-finite topological spaces

2012-06-09T10:56:36Z

We recall that a topological space $(X,\tau)$ is Pseudo-finite, if each compact subset of $X$ is finite.

One of the classical example of Pseudo-finite topological spaces can be considered as an uncountable set $X$ with the co-countable topology.(i.e.each subset with countable complement is open)

The above topology has no isolated point but it fails to be at least Hausdorff. On the base of my Knowledge there are two Tychonoff Pseudo-finite topological spaces as follows:

A. All discrete spaces are trivial examples of these spaces.

B. Consider the set $\Sigma=\mathbb{N}$$\cup$ {$p$}, and topologize it as follows:

Consider a free ultrafilter $\mathcal{U}$ on $\mathbb{N}$.
All points of $\mathbb{N}$ are isolated.
The Neighborhoods of $p$ are of the form: $U$$\cup$ {$p$}, where $U \in \mathcal{U}$.

We must recall that Case "B" is a special Example of maximal Hausdorff topologies on a set.

But I think there is no example of a Pseudo-finite Tychonoff space without isolated point !. and I guess the following statement:

Statement:Every Pseudo-finite Tychonoff space has an isolated point.

Is there a counterexample of the above statement?

Answer by AliReza Olfati for Products with compactly generated spaces

2012-06-07T13:47:09Z

I think the space $Z=\mathbb{Q}^* \times \mathbb{Q}$ works for your Question. (Here $\mathbb{Q}^*$ is the one point compactification of the rational numbers)

At first let me recall some properties of $\mathbb{Q}^*$ which you could find Them in the Monthly Article (Between $T_1$ and $T_2$) http://www.jstor.org/discover/10.2307/2316017?uid=3738280&uid=2&uid=4&sid=47699073521117.

$\mathbb{Q}^*$ is a $KC$ space.(i.e. every compact subset of this space is closed)
It's easy to show that $\mathbb{Q}^*$ and $\mathbb{Q}$ are compactly generated.

Now I bring Two theorems from the Article "On KC and K-spaces" which are important in the sequel to show that $Z$ is the space which you needed. you could find this Article from:

http://texedores.matem.unam.mx/publicaciones/index.php?option=com_remository&Itemid=57&func=startdown&id=304

Theorem 1: Let $X , Y$ be topological spaces.If $X$ is $KC$ and $Y$ is hausdorff; Then $X\times Y$ is $KC$.

Theorem 2: In the topological space $X$ the following are equivalent:

$X$ is locally compact and Hausdorff.
$X^*\times X$ is $KC$ and compactly generated.

For the sake of Theorem 1, you could see that $\mathbb{Q}^* \times \mathbb{Q}$ is $KC$. Also from part one of theorem 2 and because $\mathbb{Q}$ is not locally compact, We can conclude that $\mathbb{Q}^* \times \mathbb{Q}$ is not compactly generated.

About subspaces of $F$-spaces

2012-05-16T12:52:30Z

A topological space $X$ is an $F$-space, if Every finitely generated ideal in the ring of all continuous functions on $X$,denoted by $C(X)$, is principal. The text "Rings of continuous functions" written by Gillman and Jerison , has numerous equivalent conditions of these topological spaces, which some of them are topologic.(see pages 205-215 of the mentioned refrence). Here is an important topological equivalence of these spaces:

Def: A subspace $A$ of topological space $X$ is $C^*$-embedded if every bounded continuous real valued functions on $A$, can be extended to all of $X$.

Theorem:A topological space $X$ is an $F$-space iff every cozero-set$($i.e. $X-Z(f)$ for some $f\in C(X)$$)$ is $C^*$-embedded in $X$.

The mentioned refrence, tells us that every $C^*$-embedded subset of an $F$-space is an $F$-space.since cozero-sets and countable subsets of these spaces are $C^ *$-embedded and then $F$-space.

With the above summary I can give my questions.

Q1. The refrence didn't mention to an $F$-space that has a subspace which is not $F$-space. is this space very simple to find?

Q2. Is there a topological space $X$ with the property that every countable subspace is $C^ *$-embedded, but $X$ not an $F$-space?

Minimal prime ideals and Axiom of Choice(revised version)

2012-06-03T19:24:48Z

From the page:

http://mathoverflow.net/questions/98549/existence-of-prime-ideals-and-axiom-of-choice,

I have found that The existence of prime ideals in commutative rings is equivalent to the Boolean Prime Ideal theorem. But $BPI$ is weaker than Axiom of choice. this means that The existence of prime ideal in commutative rings with unity is weaker than $AC$. Know Another Question came in my mind that I think It is a bit different from that one. Let me recall the following theorem:

Theorem:For any commutative unitary ring $R$ there exists a minimal prime ideal.

To proving this result One can pickup a prime ideal, and throw it in a maximal chain of prime ideals(Zorn's lemma) and then the intersection of this chain gives a minimal prime ideal at hand.

You Know that the existence of minimal prime ideal needs to apply one of the equivalences of $AC$ (i.e.Zorn's Lemma) But I didn't see anything about the converse of Above theorem.

STATEMENT:Is it true that The existence of minimal prime ideals in commutative unitary rings is equivalent to $AC$.

I am interested in To Know if the situation changes When we give minimality Condition on Prime ideals.

I think its better to recall the difference of two following situations in topology:

The statement "product of compact Hausdorff spaces is compact", does not implies $AC$

But

The statement "product of compact spaces is compact" is equivalent to $AC$

Existence of prime ideals and Axiom of Choice.

2012-06-01T08:22:43Z

One of the must obvious equivalences of Axiom of Choice is the converse of Krull Theorem. Bernhard Banaschewski in the Article titled by A New Proof that “Krull implies Zorn” showed a very simple proof of the following statement.

Theorem: the existence of maximal ideals in a ring with unity is equivalent to Axiom of choice.

This means that every attempt to prove the existence of maximal ideals is related to apply the Axiom of Choice.

Another important theorem in commutative algebra is Cohen's theorem, which tells us that if $R$ is a commutative ring with unity and $I$ is an ideal of $R$ disjoint from a multiplicative closed subset $S\subset R$, then there exists a prime ideal $P$ so that $I \subset P$ and $P\cap S=\varnothing$.

Cohen's theorem implies that In a commutative ring with unity there exists a prime ideal. Notice that this prime ideal need not be a maximal ideal but we need to apply Zorn's Lemma to show the existence of it. Now Here are my Questions:

Is it true that For Showing the existence of prime ideal in a commutative ring with unity we need the Axiom of choice or we can show the existence of it without applying this Axiom?
If the Answer of above Question is negative, what kind of Axiom weaker than Axiom of choice is needed to show the existence of prime ideals in a commutative ring?
What kind of relation is between the Axiom of countable choice and The existence of prime ideals in a commutative ring with unity?

Basis for completely regular spaces(Tychonoff Spaces)

2012-04-29T19:08:26Z

If the space $X$ is completely regular, we Know that The collection {$intZ(f)$:$f$ is a continuous function from $X$ to the real numbers} is an open base for open subsets of the space $X$ (i.e. If for each element $x$ and each open set $U_x$ of $X$, there exist a continuous real valued function $f:X→\mathbb{R}$ such that $x∈intZ(f)⊆Z(f)⊆U_x)$. I have two questions about converse of this theorem. these questions are almost the same, but I think these are different.

1.If for each element $x$ and each open set $U_x$ of $X$, there exist a continuous real valued function $f:X→\mathbb{R}$ such that $x∈intZ(f)⊆U_x$, then $X$ is completely regular.

2.If for each element $x$ and each open set $U_x$ of $X$, there exist a continuous real valued function $f:X→\mathbb{R}$ such that $x∈intZ(f)⊆Z(f)⊆U_x$, then $X$ is completely regular.

I think these two claims have counterexamples and these conditions don't emply the complete regularity of $X$.

Lindelöf subsets of $P$-spaces

2012-05-22T11:00:51Z

A completely regular topological space $(X,\tau)$ is called a $P$-space, if every $G_\delta$-subset of $X$ is open .$($i.e $\tau$ is closed under countable intersection$)$. Here we recall some special properties of $P$-spaces:

Every countable subset of $X$ is obviously closed and discrete.
Every countable subset of $X$ is $C$-embedded in $X$.$($i.e. every continuous real valued function on a countable subset of $X$ can be extended to all of $X$ $)$

Now with the sake of above properties I could pose my Questions. My questions that are given as follows are the extended form of these properties of countable sets to Lindelöf subsets of $P$-spaces.

Is it true that in every $P$-space, every Lindelöf subset is closed?
Is it true that in every $P$-space every Lindelöf subset is $C$-embedded in $X$?

Answer by AliReza Olfati for Rings in which every non-unit is a zero divisor

2012-05-17T19:15:19Z

I think if you have some information about rings of continuous functions$(C(X))$, you can construct a wide class of rings with this property.

At first let me give you some special information about these examples.

Def 1. For topological space $X$ we denote the ring of all continuous functions on $X$ by $C(X)$. for $f\in C(X)$ the zero-set of $f$ is defined as: $Z(f)=${$x\in X$: $f(x)=0$}

Def 2. A completely regular topological space $X$ is called an almost $P$-space, if for every $f\in C(X)$ with nonempty zero-set ,i.e.$Z(f)$, this set has nonempty interior, i.e.there exist $x$ that $x\in int_X Z(f)$.

With the above definition, I can introduce a theorem which classifies all Rings of continuous functions $C(X)$ with the property that every non-unit is a zero-divisor.

Theorem: In the ring $C(X)$, every non-unit is a zero divisor iff the topological space $X$ is almost $P$-space.

The simplest examples of almost $P$-spaces are discrete spaces. for example if $X$ is a discrete space, then $C(X)$ is equal to the usual cartesian product $\mathbb{R}^X$. So for arbitrary set $X$ you can construct $\mathbb{R}^X$ to have the property of your question.

Comment by AliReza Olfati

2012-11-27T07:34:04Z

Dear all, I think if $R$ is an integral domain, in the ring $M_2(R)$ your statement is trivially true. because in this case, we could characterize all of idempotents. all of idempotents have the form $0$, $I_2$ or a matrix whose entries are $a_{11}=a$, $a_{12}=b$ , $a_{21}=c$, $a_{22}=1-a$ and $a,b,c$ satisfy in the relation $bc=a-a^2$. so in this case the only nonzero ideal in $M_2(R)$ which contains an idempotent is the total ring. so in this case, your claim is true.

Comment by AliReza Olfati

2012-11-26T06:28:45Z

Dear Alex. Please check it more precisely. You could not change the intervals in x-axis to them. please look at the basis of $(0,0)$ and $(1,0)$. each of neighborhoods of $(0,0)$ geometrically should contains the rectangular rigion which has the fixed length equal to $\frac{1}{2}$. also each of neighborhoods of $(1,0)$ contains the rectangular rigion which has the fixed length equal to $\frac{1}{2}$. so you could not change them in your favor.

Comment by AliReza Olfati

2012-11-26T05:59:08Z

Hi.Let $F$ be an infinite field, then there exist infinitely many distinct pair $(I,J)$ of minimal left ideals of $M_2(F)$ such that $M_2(F)=I \oplus J$. so this shows that in this case $F$ has only two trivial idempotent, But $M_2(F)$ has infinitely many nontrivial idempotent. you could find this point at exercise [11.b] page 443 of Hubgerford. So I think you should determine the property of your ring$R$ which entries of matrix ring come from it.

Comment by AliReza Olfati

2012-11-23T04:31:14Z

@ Goldstern, is this space hausdorff and zero- dimensional? for zero-dimensionality of a topological space, I mean it has a base of clopen subsets.

Comment by AliReza Olfati

2012-11-22T19:51:57Z

Dear Goldstern, I should recall that a P-space is compact if and only if it is finite. so it doesn't make sense in our problem. The reference for the well-Known theorem in my question is: exercise [12 H.6] of chapter 12 in the text "rings of continuous functions".

Comment by AliReza Olfati

2012-11-22T17:23:53Z

@Asaf. you are right. I am not a set Theorist and since i am working in ZFC, i have thought that this problem is an open problem. I do not Know why I have thought that this is an unsettled problem. But there is another Question. you claim that this is unprovable. when we consider this, can we assume if there is a least measurable cardinal or there is not a measurable cardinal, if we work in "ZFC"? for notation i will fixe it as soon as possible.

Comment by AliReza Olfati

2012-11-22T12:16:36Z

@ Asaf, I only assumed that "if" this cardinal exists. In this case it's not important for me to work in which Axiomatic system larger than "ZFC". for unsettled i mean unprovable in "ZFC" to

Comment by AliReza Olfati

2012-11-11T11:50:37Z

Thanks dear Alborz. I think you are skilled in the existence of maximal subrings. could you think about the second Question.If you are familiar with the kolmogoroff theorem in characterization of maximal ideals of C(X), when X is compact, You could find that it's a bit similar to it. cheers

Comment by AliReza Olfati

2012-10-20T13:22:20Z

It is easy. because $X$ is dense in $Y$. We Know that if $f:Y \rightarrow Z$ and $Z$ is hausdorff, and $f,g$ are the same on a dense set then $f=g$

Comment by AliReza Olfati

2012-10-20T13:12:12Z

You bring the contents of the text Rings Of continuous Functions to This site. Please look at that book more precisely other than Pose them as a Question. They are simple home works which are not appropriate for this site.

Comment by AliReza Olfati

2012-09-18T16:43:29Z

Ah. Thenk you very much for your notifications. I am sorry for duplication of this Question. I didn't found it in MO.

Comment by AliReza Olfati

2012-08-29T11:09:15Z

Dear jbc. Thank you so much for your nice and simple construction, from your answer for part 1 of my Question, we could deduce that if the condition of part 2 occurs, then $X$ should be real compact. but it also remains to approach to the compactness of it. Best wishes

Comment by AliReza Olfati

2012-08-28T20:21:49Z

Dear Josh. Thank you very much for your notation. but it's better to see that in the study of rings of continuous functions, we could suppose that the topological space that all continuous functions defined on it, is completely regular Hausdorff space. then the case which you mentioned as an answer could not occur. As you Know for every topological space $X$ there exists a completely regular Hausdorff space $Y$ so that $C(X)$ and $C(Y)$ are ring isomorphic.

Comment by AliReza Olfati

2012-08-28T20:12:40Z

@Ollie, You have said that for the chain $(R_i)_{i\in I}$, $R=\cup_{i\in I} R_i$ is also proper. Why do you sure that this subring is proper? I think one of the conditions for applying the Zorn argument fails and that is being proper of the mentioned subring.

Comment by AliReza Olfati

2012-08-28T17:41:23Z

Dear Ollie, My notation of $C(X)$ consider all continuous real valued functions. On the other hand I do not understand How you could apply Zorn's lemma for this situation. Please describe in more details. Thanks a lot.Also How could you deduce that $\mathbb {C}= C(\{\emptyset \})$ contains a proper maximal subring.