2
$\begingroup$

$(X, \tau_X) $ and $(Y, \tau_Y) $ be two topological spaces.

$\forall f\in Y^X$ with $\text{Gr}(f) $ is closed implies $f\in C(X, Y) $.

Question : Does this implies $(Y, \tau_Y) $ is compact?


Notation:

$Y^X$: Set of all functions from $X$ to $Y$.

$C(X, Y) =\{f\in Y^X: f \text{ is continuous }\}$

$\text{Gr}(f) =\{(x,f(x)):x\in X\}\subset X×Y$


Question $1$:(prove/disprove)

$(\forall X$ and $\forall f\in Y^X \text{ with } \textrm{Gr}(f) \subset X\times Y$ closed $\implies f\in C(X, Y) ) \implies Y$ is compact.

Question $2$:(prove/disprove)

For a fixed non discrete space $(X, \tau) $ and $\forall f\in Y^X$ having closed graph is continuous then $Y$ is compact.

$\endgroup$
1
  • 1
    $\begingroup$ You change your question in the very last sentence. Of course you need to exclude $X$ discrete for this to have a chance of being true (I don't know whether or not it is, though I doubt that it is true for every non-discrete space $X$), and you say so at the end, but it should be part of the initial question. Or maybe your question is whether there is some $X$ for which the implication holds (for all $Y$), or whether the hypothesis for all $X$ implies that $Y$ is compact? $\endgroup$
    – LSpice
    Sep 1 at 3:31

1 Answer 1

5
$\begingroup$

Let me try to answer these questions under the assumption that $Y$ is $T_1$.

We start with a set $E$ along with a filter $\mathcal F\in\mathcal P(E)$. We can then cook up a topological space $X$ with underlying set $\{x\}\sqcup E$ with topology given by the discrete topology $E^\delta$ on $E$, and $\{x\}\sqcup U$ for every $U\in\mathcal F$. It follows from definition that, for every topological space $Y$, a map $f\colon X\to Y$ is continuous if and only if $f\rvert_E\colon E\to Y$ tends to $f(x)\in Y$ along the filter $\mathscr F$, namely, for every neighborhood $V$ of $f(x)\in Y$, there exists a set $U\in\mathcal F$ such that $f(U)\subseteq V$. In short, the continuity captures the convergence along the filter.

What about the graph of a map $f\colon X\to Y$ being closed? Here we need the assumption of $Y$ being $T_1$, so that we only have to test the existence of open neighborhoods at every point $(x,y)\in X\times Y$ where $y$ runs through all points of $Y\setminus f(x)$. Unwinding the definitions, it is equivalent to saying that, there exists a subset $U\in\mathcal F$ and an open neighborhood $V$ of $y\in Y$ such that

  1. $y\not\in V$; and
  2. for every $x\in U$, we have $f(x)\not\in V$.

Equivalently, it is saying that $y$ is not a cluster point of $f$ along $\mathcal F$.


A positive answer to Question 1. It is known that, a topological space is compact if and only if every filter on the underlying set admits a cluster point. We assume that $Y$ is non-compact, thus it is non-empty, and there exists a filter on $Y$ which does not admit any cluster point. We fix such a filter $\mathcal F$, an arbitrary point $y\in Y$, and let $E=Y$ be a set endowed with the filter $\mathcal F$. We now define $f\colon X\to Y$ by $f(x)=y$ and $f\rvert_E=\operatorname{id}$. The previous analysis shows that $f$ is not continuous but its graph is closed.


A sketch of a negative answer to the second question. More precisely, we have

Proposition. For every topological space $X$, there exists a non-compact but Hausdorff topological space $Y$ such that every map $X\to Y$ with closed graph is continuous.

Roughly speaking, the compactness of $Y$ can be more complicated than what a fixed topological space $X$ can see. To see this, we introduce a slightly "quantitative" version of compactness:

Definition. Let $\kappa$ be a strong limit cardinal (i.e., for every $\lambda<\kappa$, we have $2^\lambda<\kappa$). We say that a topological space $X$ is $\kappa$-compact if, for every set $E$ with $\lvert E\rvert<\kappa$ and every filter $\mathcal F$ in $E$ (note: $\kappa$ being strong limit implies that $\lvert\mathcal F\rvert\le 2^{\lvert E\rvert}<\kappa$), every map $E\to X$ admits a cluster point along $\mathcal F$.


Update: This definition is equivalent, by the axiom of choice, to the following simpler one: every filter base $\mathcal F$ on $X$ with $\lvert\mathcal F\rvert<\kappa$ admits a cluster point in $X$. The usual proof shows that this $\kappa$-compactness is equivalent to the open-cover condition that every open cover of size $<\kappa$ contains a finite subcover.


Lemma. Let $\kappa$ be a strong limit cardinal and $Y$ a $\kappa$-compact topological space. Then for every topological space $X$ with $\lvert X\rvert<\kappa$, the projection $X\times Y\to X$ is closed.

The usual proof leads to

Corollary. Let $\kappa$ be a strong limit cardinal and $Y$ a $\kappa$-compact topological space. Then for every topological space $X$ with $\lvert X\rvert<\kappa$, a map $X\to Y$ is continuous if and only if its graph is closed.

Now in order to see the Proposition above, it suffices to construct a $\kappa$-compact non-compact topological space for a chosen strong limit cardinal $\kappa>\lvert X\rvert$. If I am not mistaken, we can take an ordinal $\lambda$ with cofinality greater than $\kappa$, with the order topology.

$\endgroup$

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.