Is $X$ compact under these conditions?

Question

Suppose $X$ is a normal space so that $\forall$ topological spaces $Y$ if $f:X \to Y$ is an embedding, then $f(X)$ is closed in $Y$. Does it follow that $X$ is compact?

Answer 1

Restating this so that $Y$ is quantified:

Suppose $X$ is a normal space such that for every topological space $Y$ and every imbedding $f:X\to Y$, the image $f(X)$ is closed in $Y$. Then $X$ is compact.

Choose $Y$ to be a compact space. Since $f(X)$ is closed in $Y$, it is compact, and as $f:X\to f(X)$ is a homeomorphism, $X$ must be compact.

There's a flaw in this argument, see if you can find it and fix it. Finished answer below.

This assumes there exists a compact space $Y$ with at least one imbedding $f:X\to Y$. We need to exhibit such a space to complete the argument. One such space is the Stone-Cech compactification of $X$, which Crostul mentions in the comments.

Answer 2

The space $X$ is Tychonoff (compeletely regular, $T_1$; follows from normal, if normal also includes $T_1$), so embedds via $e: X \to [0,1]^I$ for some set $I$ (Tychonoff's embedding theorem, the universal space fact on this page), and so by assumption, also for this particular embedding, $e[X]$ is closed in $[0,1]^I$ and hence compact, and so is the homeomorphic $X$.

In general a Hausdorff space $X$ is called $H$-closed if for every embedding $f:X \to Y$ where $Y$ is Hausdorff, $f[X]$ is closed in $Y$.

There exist $H$-closed spaces that are not compact. And a $T_3$ $H$-closed space must be compact. All this is classical (though not treated in many text books).

So the answer is easily yes if your normal includes $T_1$. Otherwise:

I call a space $X$ universally closed if for all spaces $Y$ and all embeddings $f: X \to Y$ we have that $f[X]$ is closed in $Y$. (Note that this is like $H$-closed but without the Hausdorff requirements.)

Fact: $X$ is universally closed iff $X=\emptyset$. Proof: $X=\emptyset$ is indeed universally closed as the empty set is closed in any topology. And if $X \neq \emptyset$, define $Y^\ast = X \cup \{\infty\}$ for some new point $\infty \notin X$ and let its topology be $\mathcal{T}_X$ (the topology of $X$) together with $Y^\ast$. The inclusion map $f: X \to Y^\ast, f(x)=x$ is an embedding (trivial) and $f[X]$ is not closed in $Y$ as $\{\infty\}$ is not open in $Y^\ast$ (or $\infty \in \overline{X}\setminus X$). (Note that $Y^\ast$ is even compact and normal if $X$ is.) So $X$ is not universally closed.

So in short, without some condition on $X$ or the spacs we can embed in for testing closedness, the notion is literally void.

Summary: if normal is just about separating two disjoint closed sets, then the only space that (is normal and) obeys the closed embedding property is the empty space (which is compact). If normal does imply Hausdorff we can apply the Tychonoff cube (or Cech-Stone compactification) embedding and $X$ is compact that way, and we get the class of all compact Hausdorff spaces, which is a bit more useful than the empty space alone..