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Suppose that $(X,\tau)$ is a locally compact, but not compact, Hausdorff topological space and $f:X\to\mathbb{C}$ is a continuous function. A sequence $(x_n)_{n\in\mathbb{Z}_+}$ in $X$ is said to have property (P) if, whenever $U\subseteq X$ and there exists some $A\subseteq U$ such that $A^c$ is compact, then there must exist some $N\in\mathbb{Z}_+$ such that for all $n>N$, $x_n\in U$. Intuitively, property (P) requires that the sequence eventually escape from all compact subsets of $X$ (or “diverge to infinity,” if you will).

Suppose that $f$ has the following property: If $(x_n)_{n\in\mathbb{Z}_+}$ is a sequence that has property (P), then for any $\varepsilon>0$ there exists some $N\in\mathbb{Z}_+$ such that for all $n>N$, $|f(x_n)|<\varepsilon$.

$\textbf{Claim:}\quad$ If $f$ has the said property, then the set $$T_{\varepsilon}\equiv\{x\in X:|f(x)|\geq\varepsilon\}$$ must be compact for all $\varepsilon>0$.

I tried many ways to prove this claim, but I am still struggling. I have already proved the converse: if $f$ is continuous and the $T_{\varepsilon}$ are compact, then $(f(x_n))$ converges to $0$ whenever $(x_n)$ has property (P). The other direction has eluded me so far. I wonder if you could give me some hints. Thank you in advance.


$\textbf{Update:}\quad$ I realized that the statement may not be true. I conjecture that if $X$ has a really nasty structure in terms of countability (in particular, if it is not even $\sigma$-compact), then there may exist no sequence with property (P). There may be so many compact subsets of $X$ that no countable sequence can eventually escape from all of them. (But note that this is merely a conjecture.) In this case, any continuous function vacuously has the property in the main claim (because there are no sequences with property (P)), even those for which not all of the $T_{\varepsilon}$ are compact.

In turn, I reformulated the statement in terms of nets, stating that if $f$ is such that for any net $(x_{\alpha})_{\alpha\in A}$ ($A$ is a nonempty directed index set) that eventually escapes from any compact subset of $A$ the associated net $(f(x_{\alpha}))$ in $\mathbb{C}$ tends to $0$, then the $T_{\varepsilon}$ must be all compact. In this case, the proof follows in both directions.

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It seems the following.

Your conjecture is right, such a nasty space $X$ exists. For instance, let $X$ be $\omega_1$ endowed with the order topology. Then $X$ is a locally compact countably compact non-compact space and each sequence $S$ in $X$ is bounbed by a countable ordinal $\alpha$ and hence $S$ is contained in a compact segment $[0;\alpha]\subset\omega_1$. Then any continuous function vacuously has the property in the main claim (because there are no sequences with property (P)), and for a constant function $f\equiv 1$ the set $T_{\frac 12}=X$ is not compact.

From the other side, maybe we can characterize all locally compact non-compact spaces satisfying the statement. For instance, maybe the statement holds for the space $X$ iff $X$ is $\sigma$-compact (I did not think about it).

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  • $\begingroup$ Great answer, thank you, @AlexRavsky! Upon reflection, I think I am comfortable with the modified result with nets. You know, I just started to delve into topology only recently, and when I think about “convergence,” it is always sequences that first come to my mind. But as topologies begin to lose resemblance to the standard Euclidean one (as the set of countable ordinals with the order topology in your example), it is nets, as opposed to sequences (which, of course, are just very special nets), that capture the notion of convergence in a more suitable way. $\endgroup$
    – triple_sec
    Oct 26, 2013 at 16:16
  • $\begingroup$ For example, between any two topological spaces, a function is continuous at a point if and only for each net in the domain space converging to the input point, the associated net of function values in the image space converges to the output point. In this case, too, the two-way result is true only for nets and not necessarily for sequences. Analogously, I am happy with the result holding for nets in the present context. $\endgroup$
    – triple_sec
    Oct 26, 2013 at 16:18

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