5
$\begingroup$

Let $(X,\mathcal{T})$ be a locally compact separable Hausdorff space and $A \in \mathcal{T}$ open. Does there exist a sequence $(f_n)_{n \in \mathbb{N}}$ of (bounded) continuous functions such that $$f_n(x) \uparrow 1_A(x)$$ for all $x \in X$? It is well-known that this result holds if

  • $(X,d)$ is a metric space (proof)
  • $X$ is locally compact with a countable basis (because then it is metrizable).

I thought about applying Urysohn's lemma, but actually this leaves me with finding a sequence $(K_n)_n$ of compact sets such that $K_n \uparrow A$. In the meantime I recognized that - in general - there does not exist such a sequence (see this question). Any suggestions?

$\endgroup$
5
$\begingroup$

It seems the following.

The answer is negative. Let $I=[0;1]$ be the unit segment endowed with the standard topology. Put $X=I^{\omega_1}$. By Tychonoff Theorem $X$ is compact. By Hewitt-Marczewski-Pondiczeri Theorem [Eng, 2.3.15], $X$ is separable. For each subset $S$ of the set $\omega_1$ as $\pi_S$ we denote the projection from $X=\prod\{I_\alpha\colon \alpha\in\omega_1 \}\to\prod\{I_\alpha\colon \alpha\in S\}$. Put $A=\bigcup_{\alpha\in\omega_1}\pi_{\{\alpha\}}^{-1}([0;1/2))$. Then $A$ is an open subset of the space $X$. Suppose that a function $1_A$ is a pointwise limit of a sequence $(f_n)$ of continuous functions. By Theorem, each $f_n$ is depending only of countable number of coordinates [Eng, Problem 2.7.12 (d)], that is there exists a countable subset $S_n$ of $\omega_1$ and a function $g_n: \prod\{I_\alpha\colon \alpha\in S_n\}\to I$ such that $f_n=g_n\pi_{S_n}$. Then the function $1_A$ is depending only of countable number of coordinates (from a family $\bigcup S_n$), a contradiction.

Reference

[Eng] Ryszard Engelking, General Topology, 2nd ed., Heldermann, Berlin, 1989.

$\endgroup$
  • $\begingroup$ What is $\omega_1$? $\endgroup$ – saz Feb 28 '14 at 7:19
  • $\begingroup$ @saz The first uncountable ordinal. You can read $\frak c$ instead of it. $\endgroup$ – Alex Ravsky Feb 28 '14 at 9:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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