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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?

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Since we are dealing with upper and lower semicontinuous functions instead of continuous functions, it is fruitful to consider all topological spaces instead of just completely regular spaces. The following theorem characterizes all such spaces, and $2\rightarrow 1$ incorporates François Dorais's idea in his answer to the previous question. As before, the answer involves the first measurable cardinal, and all such $T_{1}$-spaces are discrete if there does not exist a measurable cardinal. Since a space has arbitrarily small lower semicontinuous functions if and only if it has arbitrarily large upper semicontinuous functions, it suffices to characterize the spaces with arbitrarily large upper semicontinuous functions.

Theorem: Let $X$ be a topological space. Then the following are equivalent.

  1. For every $x\in X$, the neighborhood filter $\mathcal{N}(x)$ of $x$ is the intersection of finitely many $\sigma$-complete ultrafilters.

  2. For every countable partition $P$ of $X$ there is an open cover $\mathcal{U}$ of $X$ such that for each $U\in\mathcal{U}$ there are $A_{1},...,A_{n}\in P$ with $U\subseteq A_{1}\cup...\cup A_{n}$.

  3. For every countable partition $P$ of $X$ there is a countable open cover $\mathcal{U}$ such that for each $U\in\mathcal{U}$ there are $A_{1},...,A_{n}\in P$ with $U\subseteq A_{1}\cup...\cup A_{n}$.

  4. If $\epsilon_{x}\in\mathbb{R}$ for all $x\in X$, then there is an upper semicontinuous function $f:X\rightarrow\mathbb{R}$ with $f(x)\geq\epsilon_{x}$ for $x\in X$.

  5. If $n_{x}\in\mathbb{N}$ for all $x\in X$, then there is an upper semicontinuous function $f:X\rightarrow\mathbb{N}$ with $f(x)\geq n_{x}$ for $x\in X$.

Proof:

$1\rightarrow 2$. Assume that every neighborhood filter $\mathcal{N}(x)$ is the intersection of finitely many $\sigma$-complete ultrafilters. Let $x\in X$ and assume that $\mathcal{N}(x)=\mathcal{M}_{1}\cap...\cap\mathcal{M}_{n}$ where $\mathcal{M}_{1},...,\mathcal{M}_{n}$ are $\sigma$-complete ultrafilters on $X$. Let $P$ be a countable partition of $X$. Then there are $R_{1},...,R_{n}\in P$ where $R_{i}\in\mathcal{M}_{i}$ for $1\leq i\leq n$, so $R_{1}\cup...\cup R_{n}\in\mathcal{M}_{1}\cap...\cap\mathcal{M}_{n} =\mathcal{N}(x)$, so if $U_{x}=(R_{1}\cup...\cup R_{n})^{\circ}$, then $\{U_{x}|x\in X\}$ is the required open cover of $X$.

$2\rightarrow 5$ Assume $n_{x}\in\mathbb{N}$ for $x\in X$, and let $A_{n}=\{x\in X|n_{x}=n\}$. Then $\{A_{n}|n\in\mathbb{N}\}$ is a countable partition of $X$. Define a function $f:X\rightarrow\mathbb{N}$ where if $x\in X$, then $f(x)$ is the smallest natural number such that $x\in(A_{1}\cup...\cup A_{f(x)})^{\circ}$. Then one can see that $f(y)\leq f(x)$ whenever $y\in(A_{1}\cup...\cup A_{f(x)})^{\circ}$. Therefore the function $f$ is upper semicontinuous. Furthermore, we must have $f(x)\geq n_{x}$. Therefore, $f$ is the required function.

$5\rightarrow 4$. This is trivial.

$4\rightarrow 3$. Assume that $P=\{A_{n}|n\in\mathbb{N}\}$ is a countable partition of $X$. Let $n_{x}=n$ whenever $x\in A_{n}$. Then there is an upper semicontinuous function $f:X\rightarrow\mathbb{R}$ with $f(x)\geq n_{x}$ for all $x\in X$. Then $\{f^{-1}(-\infty,n)|n\in\mathbb{N}\}$ is a countable open cover of $X$. However, if $n\in\mathbb{N}$ and $x\in f^{-1}(-\infty,n)$, then $n_{x}\leq f(x)<n$, so $x\in A_{n_{x}}\subseteq A_{0}\cup...\cup A_{n-1}$. Therefore $f^{-1}(-\infty,n)\subseteq A_{0}\cup...\cup A_{n-1}$.

$3\rightarrow 2$. This is trivial.

$2\rightarrow 1$. Let $x\in X$. We claim that the neighborhood filter $\mathcal{N}(x)$ is $\sigma$-complete. Assume that $\{A_{n}|n\in\mathbb{N}\}$ is a descending sequence of neighborhoods of $x$. Let $P=\{A_{0},A_{0}\setminus A_{1},...,A_{n}\setminus A_{n-1},...,\bigcap_{n}A_{n}\}$. Then since $P$ is a partition of $X$, there is some $N$ where $A_{0}^{c}\cup(A_{0}\setminus A_{1})\cup...\cup(A_{N-1}\setminus A_{N})\cup \bigcap_{n}A_{n}\in\mathcal{N}(x)$. Therefore $A_{0}^{c}\cup(A_{0}\setminus A_{1})\cup...\cup(A_{N-1}\setminus A_{N})\cup \bigcap_{n}A_{n}=A_{N}^{c}\cup\bigcap_{n}A_{n}\in\mathcal{N}(x)$, so $A_{N}\cap(A_{N}^{c}\cup\bigcap_{n}A_{n})=\bigcap_{n}A_{n}\in\mathcal{N}(x)$ as well. Therefore $\mathcal{N}(x)$ is $\sigma$-complete. We now claim that the Boolean algebra $P(X)/\mathcal{N}(x)$ is finite. Otherwise, there is a countable partition $\{a_{n}|n\in\mathbb{N}\}$ of $P(X)/\mathcal{N}(x)$. However, one can easily show that this implies there is a countable partition $\{A_{n}|n\in\mathbb{N}\}$ of $P(X)$ where $a_{n}=A_{n}/\mathcal{N}(x)$ for all $n$. However, for all natural numbers $n$, we have $(A_{1}\cup...\cup A_{n})/\mathcal{N}(x)=a_{1}\vee...\vee a_{n}\neq 1$, so $A_{1}\cup...\cup A_{n}\not\in\mathcal{N}(x)$. This is a contradiction. Therefore, $P(X)/\mathcal{N}(x)$ is finite, so $\mathcal{N}(x)$ is the intersection of finitely many $\sigma$-complete ultrafilters.

QED

If there does not exist a measurable cardinal, then these spaces are fairly trivial. In fact, these spaces are essentially the pre-ordered sets $X$ where if $x\in X$, then $\{y\in X|y\leq x\}$ is finite. The $T_{1}$ such spaces are the partially ordered sets $X$ where if $x\in X$, then $\{y\in X|y\leq x\}$ is finite. If $X$ is a preordered set, then the collection of lower sets forms a topology.

On the other hand, if we assume that there is a measurable cardinal, then these spaces are much more interesting. Therefore in the remainder of this discussion, assume that there is a least measurable cardinal $\mu$. If a space has arbitrarily small lower semicontinuous functions, then the intersection of less than $\mu$ open sets is open. In particular, every regular space with arbitrarily large upper semicontinuous functions is a $P$-space. Thus, every regular space with arbitrarily large upper semicontinuous functions is zero-dimensional.

If $\mathcal{U}_{x}$ is a $\sigma$-complete ultrafilter on $X$ for each $x\in X$ and $\mathcal{V}$ is also a $\sigma$-complete ultrafilter, then let $\sum_{x\in X}^{\mathcal{U}}\mathcal{U}_{x}$ be the subset of $P(X)$ where $R\in\sum_{x\in X}^{\mathcal{U}}\mathcal{U}_{x}$ if and only if $\{x\in X|R\in\mathcal{U}_{x}\}\in\mathcal{U}$.` Clearly $\sum_{x\in X}^{\mathcal{U}}\mathcal{U}_{x}$ is a $\sigma$-complete ultrafilter on $X$.

We can even describe spaces with arbitrarily large upper semicontinuous functions combinatorially in terms of the convergent ultrafilters as follows.

Now assume that $X$ is a set and $M_{x}$ is a collection of finitely many $\sigma$-complete ultrafilters on the set $X$ for each $x\in X$. Then call the system $(M_{x})_{x\in X}$ additive if

  1. each $M_{x}$ contains the principal ultrafilter $\{R\subseteq X|x\in R\}$ and

  2. If $\mathcal{U}\in M_{x}$ and $\mathcal{U}_{y}\in M_{y}$ for $y\in Y$, then $\sum_{y\in X}^{\mathcal{U}}\mathcal{U}_{y}\in M_{x}$ as well.

It turns out that if $X$ is a set and $M_{x}$ is a collection of finitely many $\sigma$-complete ultrafilters on $X$ for each $x\in X$, then the system $(M_{x})_{x\in X}$ is additive if and only if there is a topology on the set $X$ such that $\mathcal{N}(x)=\bigcap M_{x}$ for each $x\in X$.

We conclude that one may consider topological spaces with arbitrarily small lower semicontinuous functions as additive systems $(M_{x})_{x\in X}$ where each $M_{x}$ is a finite set of $\sigma$-complete ultrafilters on $X$. Furthermore, the Hausdorff spaces with arbitrarily small lower semicontinuous functions correspond to the additive systems $(M_{x})_{x\in X}$ where $M_{x}\cap M_{y}=\emptyset$ whenever $x,y$ are distinct points in $X$.

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Another way of restating the OP property is "any countable point-finite covering of $X$ is locally finite" (more or less 2) –  Pietro Majer Oct 8 '12 at 9:35
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