# Not closed somewhere $\Rightarrow$ not compact anywhere?

In "Principles of Mathematical Analysis" by Walter Rudin, Theorem 2.33 asserts that compactness is a universal property: meaning if $K \subset X$ is compact in $X$, then $K$ is compact in any embedding metric space. Theorem 2.34 asserts that "compact subsets of metric spaces are closed": meaning for any $K \subset X$, if $K$ is compact, then $K$ is closed relative to $X$.

Does this mean (by taking the contrapositive) that for any set $E$, if we can find just one metric space $X$ such that $E \subset X$ and $E$ is not closed relative to $X$, then $E$ is not compact anywhere? (This would make it easy to prove, for example, that no segment $(a, b)$ is compact...)

OR should the contrapositive be read: if $E$ is not closed relative to every embedding metric space $X$, then $E$ is not compact anywhere? (This would make the contrapositive relatively useless, since every set is closed relative to itself...)

Definitions: Let $X$ be a metric space with distance function $d(p, q)$. For any $p \in X$, the neighborhood $N_r(p)$ is the set $\{x \in X \,|\, d(p, x) < r\}$. Any $p \in X$ is a limit point of $E$ if $\forall r > 0$, $N_r(p) \cap E \neq \{p\}$ and $\neq \emptyset$. Any subset $E$ of $X$ is closed if it contains all of its limit points. For any subset $E$ of $X$, an open cover is a collection of sets $\{G_\alpha\}$ which are open in $X$, such that $E \subset \bigcup_\alpha G_\alpha$. $E$ is compact if every one of its open covers has a finite open subcover.

-
FYI, this is not the standard definition of "universal property." I would use "intrinsic property" instead. – Qiaochu Yuan Feb 16 '12 at 20:37

Although it might seem that way from the definitions, compactness is not a relative property. If $E\subseteq X$, $X$ a metric space, then the compactness of $E$, although phrased in terms of open coverings of $E$ by sets open in $X$, only depends on the topology of $E$ (the subspace topology, which is also the topology you get by restricting the metric $d$ to $E\times E$). Compactness is an intrinsic property.
I'm not sure if you know about subspace topologies, but any subset of a metric space can be regarded as a metric space by restricting the metric. You can show then that $E$ is compact in your sense if and only if every cover of $E$ by sets open in $E$ (for the metric topology on $E$) has a finite subcover.
So, if $E\subseteq X$ is compact, and $E\rightarrow X^\prime$ is an embedding into another metric space, meaning a homeomorphism onto its image, then the image of $E$ will be closed in $X^\prime$, because the image of $E$ will be compact, and compact subsets of metric spaces are closed. In fact, this will work for an arbitrary continuous $E\rightarrow X^\prime$. If you can find an embedding (or any continuous map whatsoever) $E\rightarrow X^\prime$ whose image is not closed, then $E$ cannot be compact.
I should add that I was confused by this when I was first learning this stuff. The definition used in the context of metric spaces makes it seem like compactness is related to the ambient metric space $X$. What cleared everything up for me was a course in general point-set topology.
"If you can find an embedding (or any continuous map whatsoever) $E \rightarrow X′$ whose image is not closed, then E cannot be compact." Thanks! This is the key question I was looking for. – jamaicanworm Feb 16 '12 at 21:01