Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $X$ ?

share|improve this question
1  
Inn general, no. Say $X$ is the real line, $\epsilon_x=x$ for $x \ne 0$ and $\epsilon_0=1$. –  Gerald Edgar Jun 19 '12 at 19:12
    
Thanks Dear Edgar. But I think I forgot to pose the important part of my Question. For this I must apologize from members of "MO". I will fixe it. –  Ali Reza Jun 19 '12 at 19:20
    
Gerald Edgar's example easily generalizes to show that the zero-set of any continuous real-valued function on $X$ would have to be open. Thus, all continuous functions from $X$ to $\mathbb R$ have to be locally constant. –  Andreas Blass Jun 19 '12 at 22:24
add comment

4 Answers

up vote 5 down vote accepted

The existence of a non-discrete T1 space with this property equivalent to the existence of a nonprincipal $\sigma$-complete ultrafilter (i.e. the existence of a measurable cardinal). Therefore, it is consistent with ZFC that all T1 spaces with that property are discrete.

First, suppose $\mathcal{U}$ is a nonprincipal $\sigma$-complete ultrafilter on the set $X$. Pick $\infty \notin X$ and define a topology on $X \cup \lbrace\infty\rbrace$ where all points of $X$ are isolated and the neighborhoods of $\infty$ are the sets $\lbrace\infty\rbrace\cup U$ where $U \in \mathcal{U}$. Suppose $\varepsilon_x \in (0,\infty)$ have been chosen for every $x \in X \cup \lbrace\infty\rbrace$. Since $\mathcal{U}$ is $\sigma$-complete, of the sets $$X_n = \lbrace x \in X : \varepsilon_x \gt 1/(n+1) \rbrace$$ eventually belong to $\mathcal{U}$. Let $n$ be such that $1/(n+1) \lt \varepsilon_\infty$ and $X_n \in \mathcal{U}$. Define $f:X\cup\lbrace\infty\rbrace\to(0,\infty)$ by $f(x) = 1/(n+1)$ when $x \in X_n \cup \lbrace\infty\rbrace$ and $f(x) = \varepsilon_x/2$ elsewhere. Then $f$ is continuous and $f(x) \lt \varepsilon_x$ for all $x \in X \cup\lbrace\infty\rbrace$.

For the converse implication, suppose $X$ is a space with the given property.

First observe that the filter generated $\mathcal{N}$ by the neighborhoods of a point $x_0 \in X$ is $\sigma$-complete. To see that $\mathcal{N}$ is $\sigma$-complete, suppose $U_0 \supseteq U_1 \supseteq \cdots$ is a sequence of open neighborhoods of $x_0$ and let $Z = \bigcap_{n\lt\omega} U_n$. If $x \notin Z$ then define $\varepsilon_x = \min\lbrace 1/(n+1) : x \in U_n\rbrace$ and define $\varepsilon_x = 1$ on $Z$ (say). Suppose $f:X \to (0,\infty)$ is continuous and pick $n \geq 1$ so that $f(x_0) \geq 1/n$. Then there is an open neighborhood $U$ of $x_0$ such that $f(x) \gt 1/(n+1)$ for all $x \in U$. Thus $f(x) \gt 1/(n+1) \geq \varepsilon_x$ for any $x \in (U \cap U_n) - Z$. So if $f(x) \lt \varepsilon_x$ for every $x \in X$, then we must have $U \cap U_n \subseteq Z$, which shows that $Z$ contains an open neighborhood of $x_0$.

If $x_0$ is not an isolated point of $X$ and $X$ is T1 then $\mathcal{F} = \lbrace N-\lbrace x_0\rbrace: N \in \mathcal{N}\rbrace$ is a free filter on $X-\lbrace x_0\rbrace$ which is also $\sigma$-complete. This is not necessarily an ultrafilter, but I will show that there is a $Y \subseteq X-\lbrace x_0 \rbrace$ such that the restriction $\mathcal{F}|Y = \lbrace A \cap Y : A \in \mathcal{F}\rbrace$ is an ultrafilter, which is necessarily also $\sigma$-complete.

Indeed, suppose for the sake of contradiction that there is no such set $Y$, then we can find a countable partition of $X-\lbrace x_0 \rbrace$ into pairwise disjoint sets $X_n$ that are not in the ideal dual to $\mathcal{F}$. (Since $\mathcal{F}$ is not an ultrafilter, we can find sets $X_0, Y_0$ that are not in the dual ideal of $\mathcal{F}$ such that $X-\lbrace x_0 \rbrace = X_0 \cup Y_0$ and $X_0 \cap Y_0 = \varnothing$. Since $\mathcal{F}|Y_0$ is not an ultrafilter, we can similarly partition $Y_0 = X_1 \cup Y_1$. Repeat ad infinitum and throw any leftover points back into $X_0$.) Given such a partition, define $\epsilon_x = 1/(n+1)$ when $x \in X_n$ and $\epsilon_{x_0} = 1$. Suppose $f:X \to (0,\infty)$ is continuous and pick $n \geq 1$ so that $f(x_0) \geq 1/n$. Then there is a neighborhood $U$ of $x_0$ such that $f(x) \gt 1/(n+1)$ for all $x \in U$. Then $f(x) \gt 1/(n+1) = \epsilon_x$ for any $x \in U \cap X_n$. Since $X_n \cap U \neq \varnothing$, otherwise $X_n$ would be in the ideal dual to $\mathcal{F}$, we conclude that $f(x) \gt \varepsilon_x$ for some $x \in X$. Thus, we contradict the fact that our space has the given property.

share|improve this answer
    
I had missed the Tychonoff requirement in my first reading of the question. I should point out that the space constructed in the second paragraph is indeed Tychonoff. For the converse implication, it makes no difference whether the space is Tychonoff or not. –  François G. Dorais Jun 20 '12 at 14:08
    
The space you constructed above in the second paragraph is the a space $Y$ where $X\subseteq Y\subseteq\upsilon X$ where $|Y\subseteq X|=1$ and where $\upsilon X$ is the Hewitt-realcompactification of $X$. –  Joseph Van Name Jun 20 '12 at 15:48
    
Hello Dear Dorais. Your Proof is Fantastic. You could relate This Problem with existence of measurable cardinal beautifully. Thank you very much for your nice description. Then with your positive answer we could find that we could not look for this Property in the spaces such as $\mathbb{N}$,$\mathbb{Q}$,$[0,\Omega_1)$ and another well Known examples with nonmeasurable cardinal. –  Ali Reza Jun 21 '12 at 6:51
    
It should be noted that a slight modification of the above proof shows that whenever a space has this property, then the neighborhood filter around every point is the intersection of finitely many $\sigma$-complete ultrafilters. In particular, if $\mu$ is the first measurable cardinal, then the intersection of less than $\mu$ open sets in open. –  Joseph Van Name Jun 27 '12 at 4:23
    
Joseph, indeed it seems to me that all such spaces are in some way "generated" by spaces like the one I gave as an example. However, I can't formulate a precise meaning for "generated" in this context (so I can't prove it either). If anyone has any thoughts to make this precise, let me know... –  François G. Dorais Jun 27 '12 at 16:48
show 1 more comment

The existence of a continuous function $f:X\rightarrow \mathbb{R} _ +$ with $ f(x) < \epsilon(x) $, for any given $\epsilon: X\rightarrow \mathbb{R} _ +$ is quite a strong assumption on $X$, as already observed in Gerald Edgar's and Andreas Blass' comments.

If we accept more reasonable assumptions on $\epsilon$, it is worth recalling the following simple but useful theorem by C.H.Dowker (see Dugundji, Topology, VIII.4.3):

If $X$ is paracompact and $\delta$ and $ > \epsilon $ are real-valued functions on $X$, $\delta$ upper semicontinuous and $\epsilon$ lower semicontinuous, and if $\delta(x) < > \epsilon (x) $ for all $x\in X$, then there exists a continuous $f$ on $X$ with $\delta(x) < f(x) < \epsilon(x)$ on $X$.

share|improve this answer
add comment

We shall characterize those spaces in terms of a partition relation. This characterization is very similar to the property given in the question, but it may be useful and insightful.

Let $X$ be a Tychonoff space. Then the following are equivalent.

  1. Whenever $\epsilon_{x}>0$ there is some $x\in X$ there is some $f:X\rightarrow(0,\infty)$ with $f(x)>\epsilon_{x}$ for $x\in X$ (the proof is simpler if we replace $< $ with $>$).

  2. If $n_{x}\in\mathbb{N}$ for $n\in\mathbb{N}$, then there is a continuous mapping $f:X\rightarrow\mathbb{N}$ such that $f(x)>n_{x}$ for $x\in X$. In other words, there are arbitrarily large continuous functions from $X$ to $\mathbb{N}$.

  3. If $P$ is a partition of $X$ into countably many sets, then there is some partition $Q$ of $X$ into clopen sets such that for each $B\in Q$ there are $A_{1},\dots,A_{n}\in P$ such that $B\subseteq A_{1}\cup\dots\cup A_{n}$.

$1\rightarrow 2$. Assume that if $\epsilon_{x}>0$ for $x\in X$ then there is a continuous mapping $f:X\rightarrow(0,\infty)$ with $f(x)<\epsilon_{x}$. Then as François G. Dorais showed, the neighborhood filter $\mathcal{N}(x)$ of every point $x\in X$ is $\sigma$-complete. Therefore the space $X$ is a $P$-space. It is well known and one can easily show that a completely regular space is a $P$-space if and only if whenever $f:X\rightarrow\mathbb{R}$ is continuous, then around each point $x\in X$ there is a neighborhood $U$ of $x$ with $f''(U)=\{f(x)\}$. In other words, $P$-spaces are precisely the spaces where every continuous real-valued function is locally constant.

Now assume that $n_{x}\in\mathbb{N}$ for $x\in X$. Then there is some function $f:X\rightarrow\mathbb{R}$ such that $f(x)>n_{x}$ for $x\in X$. Let $\mathbb{R}^{d}$ be the real numbers with the discrete topology. Then since $X$ is a $P$-space, the function $f$ is locally constant, so $f$ is a continuous function from $X$ to $\mathbb{R}^{d}$. Let $g:\mathbb{R}^{d}\rightarrow\mathbb{N}$ be a function with $g(x)\geq x$ for $x\in X$. Then we have $g\circ f:X\rightarrow\mathbb{N}$ be a continuous function with $g\circ f(x)\geq f(x)>n_{x}$ for $x\in X$.

$2\rightarrow 1$ This is obvious.

$2\rightarrow 3$. Assume that $P=\{A_{1},\dots,A_{n},\dots\}$ is a partition of $X$ into countably many sets. Then if $x\in A_{n}$, then assume that $n_{x}=n$. Then there is a continuous $f:X\rightarrow\mathbb{N}$ such that $f(x)>n_{x}$ for all $x\in X$. Let $B_{n}=f_{-1}(\{n\})$ for all $n$. We claim that $B_{n}\subseteq A_{1}\cup\dots\cup A_{n}$. If $x\in B_{n}$, then $n_{x}< f(x)=n$, so $x\in A_{n_{x}}$ for some $n_{x}< n$, so $x\in A_{1}\cup\dots\cup A_{n}$. Thus $B_{n}\subseteq A_{1}\cup....\cup A_{n}$.

$3\rightarrow 2$. Assume that whenever $P$ is a countably partition of $X$, then there is countable partition $Q$ of $X$ into clopen sets where for each $B\in Q$ there are $A_{1},\dots,A_{n}\in P$ with $B\subseteq A_{1},\dots,A_{n}$. Now assume that $n_{x}\in\mathbb{N}$ for $x\in X$. Then let $A_{n}=\{x\in X|n_{x}=n\}$ for all $n$. Then there is a partition $Q=\{B_{1},\dots,B_{n},\dots\}$ of $X$ into clopen sets such that for all $n$ there is a function $g:\mathbb{N}\rightarrow\mathbb{N}$ such that $B_{n}\subseteq A_{1}\cup\dots\cup A_{g(n)}$ for all $n$. Let $f:X\rightarrow\mathbb{N}$ be the function where if $x\in B_{n}$, then $f(x)=g(n)+1$. Then since $x\in B_{n}\subseteq A_{1}\cup\dots\cup A_{g(n)}$, we have $x\in A_{i}$ for some $i\leq g(n)$, so $n_x=i\leq g(n)< g(n)+1=f(x)$. Furthermore, since each $B_{n}$ is clopen, we have $f$ be a continuous function.

share|improve this answer
    
For some reason the end of $3\rightarrow 2$ did not show up. Here is the end of the proof. $n_{x}=i\leq g(n)<g(n)+1=f(x)$. Furthermore, since each $B_{n}$ is clopen, we have $f$ be a continuous function. –  Joseph Van Name Jun 20 '12 at 16:22
    
The reason is that you need to leave a blank space after any < character, otherwise markdown will interpret it as a start of an HTML element. –  Emil Jeřábek Jun 20 '12 at 16:25
    
Also, \{ and \} won’t show up unless the backslashes are doubled. –  Emil Jeřábek Jun 20 '12 at 16:33
add comment

So you are asking: which topological spaces, besides the discrete ones, are such that for every strictly positive real function $g$ there is a strictly positive continuous real function $f$ with $ 0 < f < g $? Only some hints.

Others have already noted that there cannot be nontrivial convergent sequences.

You can note that if there are no nontrivial convergent sequences, then replace $g$ with the largest $h<g$ which takes only values of the form $1/n$, and then note that this $h$, even if not continuous, at least does not give the above noted problem (forcing a possible continuous $f$ to have value $0$). Taking the closures of the sets where $h>1/n$ one can then use Urhyson's lemma / Tietze extension theorem (when the space is normal) and so obtain a continuous $f$ with $0<f<h$.

Is there a non-discrete normal space with no nontrivial converging sequence? You can easily find the answer in any standard book on general topology which treats Stone compactification.

Added: please replace "no nontrivial convergent sequences" with the stronger "every countable subset is closed". And examples are even more exotic, but Gillman and Jerison, rings of continuous functions, should have them (and perhaps also Engelking).

Concerning TeX commands: I do not use them on purpose, but if it this the rule to always use them here, then I will in future only give answers that do not require mathematical notation, so that we all will be able to read in the way we like.

share|improve this answer
    
NN, you need to use TeX commands. I added them. –  Bill Johnson Jun 19 '12 at 23:19
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.