Does there exist a non compact subspace A of the Euclidean space such that each continuous map of A to a Hausdorff space is closed

Question

I guess not, since a closed and bounded subset of Euclidian space is compact, so A is either not closed or not bounded.
If A is not closed, for example the inclusion map: A to $\mathbb{R^n}$ is not a closed map.
But I can not find a counterexample for A is closed, what a counterexample would that be? Thanks!

Answer 1

Let's call $A \subseteq (X,d)$, where $(X,d)$ is a metric space, "functionally-closed" if every continuous $f: A \to Y$, where $Y$ is Hausdorff, is a closed map. Standard theorems teach us that $A$ compact implies $A$ functionally-closed.

$X$ being metric, so Tychonoff, has a Hausdorff compactification $c(X)$ in which $X$ densely embeds. Then the standard embedding $i_A: A \to c(X)$ is closed iff $A$ is compact. So a functionally closed subset is compact, as the OP suspected.

Answer 2

Suppose $A\subseteq \mathbb R^n$ is such that any continuous map from $A$ to any Hausdorff space is closed.

Embedd $\mathbb R^n$ in $S^n$ (via the inverse of the stereographic projection). Then the image of $A$ in $S^n$ is closed by hypothesis (because $S^n$ is Hausdorff), hence compact (because $S^n$ is compact). Thus $A$ is compact.

Answer 3

Let $d(x,y)$ be the distance between $x,y\in \Bbb R^n.$ For $p\in \Bbb R^n$ define $f_p: A\to [0,1]$ as $f_p(x)=\frac {d(x,p)}{1+d(x,p)}.$

If $A$ is unbounded, choose any $p\in\Bbb R^n.$

If $A\ne \bar A$, choose $p\in \bar A\setminus A.$