# (Dis)prove : if every function with closed graph are continuous then the target space is compact

$$(X, \tau_X)$$ 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.

• 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? Sep 1 at 3:31

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.

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.