Metric spaces, compactness and completeness

For a metric space $X$, compactness (every open cover has a finite subcover), sequential compactness, and completeness together with total boundedness are equivalent. The Heine–Borel theorem: a subset of $\RR^n$ is compact if and only if it is closed and bounded. In $\QQ$ the set $[0,2]\cap\QQ$ is closed and bounded and not compact, and in $C([0,1])$ with the supremum norm the closed unit ball is closed and bounded and not compact. The Baire category theorem: in a complete metric space, every countable intersection of dense open sets is dense.

Topological notions in a metric space (Tag D-TVKFM)

Let $(X,d)$ be a metric space, let $A\subseteq X$, and let $p\in X$.

A neighborhood of $p$ is an open set $U_p\subseteq X$ containing $p$.
An $\varepsilon$-neighborhood of $p$ is an open ball $B_\varepsilon(p) \coloneqq \theset{q\in X \suchthat d(p, q) < \varepsilon}$ for some $\varepsilon>0$.
The point $p$ is an accumulation point or a limit point of $A$ if every punctured neighborhood $U_p\setminus\theset{p}$ contains a point $q\in A$. Write $A'$ for the set of limit points of $A$ and $\overline{A} \coloneqq A\cup A'$ for the closure of $A$.
If $p\in A$ and $p$ is not a limit point of $A$, then $p$ is an isolated point of $A$.
$A$ is closed if $A' \subseteq A$, so that $A$ contains all of its limit points.
A point $p\in A$ is an interior point of $A$ if there is a neighborhood $U_p$ of $p$ with $U_p \subseteq A$.
$A$ is open if every point of $A$ is an interior point of $A$.
$A$ is perfect if $A$ is closed and $A\subseteq A'$, so that every point of $A$ is a limit point of $A$.
$A$ is bounded if there are a real number $M$ and a point $q\in X$ such that $d(p, q) < M$ for all $p\in A$.
$A$ is dense in $X$ if every point of $X$ lies in $A$ or is a limit point of $A$, that is, $X = A\cup A' = \overline{A}$.

Completeness of a metric space (Tag D-G5N6I)

A metric space $(X,d)$ is complete if every Cauchy sequence in $X$ converges to a point of $X$.

Heine--Borel theorem (Tag T-QPTHZ)

In Euclidean space, compact means exactly closed and bounded.Suggest a slogan

A subset $X\subseteq \RR^n$ is compact if and only if $X$ is closed and bounded.

Compact if and only if sequentially compact for metric spaces (Tag PR-25GM2)

For metric spaces, compactness is equivalent to every sequence having a convergent subsequence.Suggest a slogan

A metric space $(X,d)$ is compact if and only if it is sequentially compact: every sequence in $X$ has a subsequence that converges in $X$.

Characterizations of compact subsets of a metric space (Tag T-YOZX6)

For metric spaces, compactness is the same as sequential compactness and as completeness plus total boundedness.Suggest a slogan

Let $(X,d)$ be a metric space and let $E\subseteq X$, with the metric induced by $d$. The following are equivalent [@Fol13]:

$E$ is complete and totally bounded: for every $\varepsilon>0$, $E$ is covered by finitely many balls of radius $\varepsilon$.
$E$ is sequentially compact: every sequence in $E$ has a subsequence that converges to a point of $E$.
$E$ is compact: every cover of $E$ by open subsets of $X$ has a finite subcover.

If $X$ is complete, then $E$ is complete if and only if $E$ is closed in $X$. In $\RR^n$ with the Euclidean metric, $E$ is totally bounded if and only if $E$ is bounded; combined with the theorem, this gives the Heine–Borel theorem. In a general metric space, a closed bounded subset need not be compact: in an infinite set with the discrete metric $d(x,y)=1$ for $x\neq y$, the whole space is closed and bounded but not totally bounded.

Dense subsets (Tag FD-BA2WU)

Let $X$ be a topological space. A subset $A\subseteq X$ is dense in $X$ if its closure is $X$: $\cl_X(A) = X$.

When is a subset $A \subseteq X$ dense in $X$?

Nowhere dense sets (Tag D-2MJRE)

Let $X$ be a topological space. A subset $S\subseteq X$ is nowhere dense in $X$ if its closure $\overline{S}$ has empty interior, that is, no nonempty open subset of $X$ is contained in $\overline{S}$.

Equivalently, every nonempty open set $U\subseteq X$ contains a nonempty open set $V$ with $V\cap S=\emptyset$, so $S$ is not dense in any nonempty open subset of $X$. Indeed, if $\overline{S}$ has empty interior then $V\coloneqq U\setminus\overline{S}$ is nonempty and open; conversely, a nonempty open $W\subseteq\overline{S}$ contains no nonempty open set disjoint from $S$. For $X=\RR$, $S$ is nowhere dense if and only if every open interval contains an open subinterval disjoint from $S$.

$\RR$ is not a countable union of nowhere dense sets (Tag T-7FJFK)

$\RR$ is Baire, so countably many nowhere dense sets cannot cover it.Suggest a slogan

$\RR$ is a Baire space. In particular, $\RR$ is not a countable union of nowhere dense subsets of $\RR$.

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