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?


1 Answer 1


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.


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

  • $\begingroup$ What is $\omega_1$? $\endgroup$
    – saz
    Feb 28, 2014 at 7:19
  • $\begingroup$ @saz The first uncountable ordinal. You can read $\frak c$ instead of it. $\endgroup$ Feb 28, 2014 at 9:00

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