-1

Let us assume that $X$ is a Tychonoff space and that there exists a map $\psi: X \to (0,\infty)$ such that $K_R := \psi^{-1}((0,R])$ is compact for all $R > 0$. My question is: is $X$ exhaustible by compacts?

Note that if $\psi$ is bounded, then there exists an $R>0$ such that $\psi(x) \leq R$ for all $x \in X$, implying that $X = K_R$ is compact. Being compact, in turn, implies that $X$ exhaustible by compacts.

However, if $\psi$ is unbounded, while it's still clear to me that $X$ is $\sigma$-compact, it's not clear whether we would have a sequence of compacts $(K_n)_{n\geq 1}$ such that $K_n \subset \text{int}(K_{n+1})$.

----- Edit -----

We do not assume $\psi$ to be continuous, otherwise it's rather straightforward.

----- Edit 2 -----

Ultimately, I want to find a way to guarantee that, given an arbitrary compact $K \subset X$, there exists a $K_R = \psi^{-1}(0,R]$ such that $K \subset K_R$. This without assuming the continuity of $\psi$.

Oscar
  • 1,465
  • If $\psi$ is continuous (and I think you meant to assume that), then $K_R\subseteq \psi^{-1}((0, S))\subseteq \text{int}(K_S)\subseteq K_S$ for $R < S$ – Jakobian May 16 '24 at 16:32
  • No, $\psi$ is not continuous necessarily. – Oscar May 16 '24 at 16:34
  • Compact implies exhaustible by compacts by a simple argument that if $X$ is compact then you just take $K_n = X$. Your proof is over-complicated – Jakobian May 16 '24 at 16:35
  • If $\psi$ is not continuous then this is just $\sigma$-compactness again – Jakobian May 16 '24 at 16:36
  • If $X$ is $\sigma$-compact and $K_n$ is a sequence of compact sets with $K_n\subseteq K_{n+1}$ then just let $\psi(x) = n$ for $x\in K_{n+1}\setminus K_n$ and here you have your map. – Jakobian May 16 '24 at 16:37
  • Sure, I can agree that the proof for the bounded case is overcomplicated. I also see that $\sigma$-compactness is obvious. I am interested in exhaustion though. Do you see if there's a way to attain it? (Thanks for having a look btw) – Oscar May 16 '24 at 16:39
  • I can't think of anything that together with assuming $X$ is Hausdorff wouldn't let to some kind of continuity. You can assume $\psi^{-1}((0, R))$ is open for all $R > 0$ but if $X$ is Hausdorff this would imply $\psi$ is continuous – Jakobian May 16 '24 at 16:43
  • Just note that such open ended questions are outside of scope of this site – Jakobian May 16 '24 at 16:44
  • I see your point... Well, thanks for the insight anyway! I guess I need to impose some additional structure on $\psi$... I'll work on this. Thanks! – Oscar May 16 '24 at 16:46

1 Answers1

1

Assume given $\psi:X\rightarrow(0,\infty)$ such that $K_R=\psi^{-1}(0,R])$ is compact for each $R>0$. Put

$$K_n=\psi^{-1}(0,n],\qquad U_n=\psi^{-1}(0,n +1/2)$$

Then $$K_n\subseteq U_n\subseteq K_{n+1}$$ for all $n\geq1$. Each $K_n$ is compact by assumption and $X=\psi^{-1}(0,\infty)=\bigcup_{n\geq1}\psi^{-1}(0,n)=\bigcup_{n\geq1}K_n$. Since $\psi$ is continuous, each $U_n$ is open. In particular, $U_n\subseteq K_{n+1}^\circ$. This gives $$K_n\subseteq K_{n+1}^\circ.$$ Hence that $K_n$'s are an exhaustion of $X$.

Edit:

Choose an enumeration $\mathbb{Q}$ and let $\psi:\mathbb{Q}\rightarrow(0,\infty)$ be the function assigning each rational its corresponding natural number. Clearly $\psi$ is not continuous.

Since $\psi^{-1}[0,R)$ is finite for each $R>0$, it is compact. But $\mathbb{Q}$ is not exhaustible by compacts as it is not locally compact.

Tyrone
  • 17,539
  • $U_n$ is open if $\psi$ is continuous, which I do not assume. Hence, the difficulty. – Oscar May 16 '24 at 16:36
  • I believe this assumption was added after I posted this answer. I will add a counterexample in the extra case. – Tyrone May 16 '24 at 16:38
  • @Tyrone No, its just that continuity was never mentioned in the first place – Jakobian May 16 '24 at 16:39
  • There was no assumption added, I simply emphasised that it's not necessarily continuous. – Oscar May 16 '24 at 16:39
  • It is standard in many places that a "map" is a continuous function. I have added a counterexample. – Tyrone May 16 '24 at 16:41
  • @Tyrone do you reckon that assuming $X$ to be locally compact would solve my problem (see Edit 2)? Locally compactness would ensure an exhaustion. The issue then becomes showing that the exhaustion can be given by sets of the form $\psi^{-1}((0,R])$... – Oscar May 17 '24 at 16:36
  • If $\psi^{-1}(0,R]$ is compact (or at least has compact closure) for each $R>0$, then $X$ is $\sigma$-compact (this does not need continuity of $\psi$). Then $X$ is exhaustible by compacts if and only if it is locally compact according to the wikipedia link. – Tyrone May 17 '24 at 18:41
  • Regarding the edit, one kind of trivial way to get what you want is for all compact subsets of $X$ to be finite. I think the only reasonable way forwards is to put some basic assumptions on $\psi$. For instance you could assume it is lower-semicontinuous rather than continuous. There are probably other generalised continuity properties as well which might be useful there. – Tyrone May 17 '24 at 19:00
  • Thanks for the reply @Tyrone. I reckon that $\psi$ is already lower semicontinuous necessarily. This follows from the initial assumptions. I'm really trying to avoid imposing continuity on $\psi$. I'll keep trying for a bit more thinking of ways to achieve Edit 2 by imposing some more structure on $X$. Thanks! – Oscar May 18 '24 at 11:08