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:


$$\forall x \in X $$ 


$$0< f(x) < \epsilon\_x$$




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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 $X$ ?



Existence of an arbitrary Small positive continuous real Valued Function
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.


