Let
$$X_1\subseteq X_1\subseteq X_2\subseteq\dots$$
be an expanding sequence of closed embeddings of separable metric spaces. Let $X$ be the colimit of the sequence. We can assume that $X=\bigcup_\mathbb{N}X_n$ with the topology in which a subset $U\subseteq X$ is open if and only if $U\cap X_n$ is open in $X_n$ for each $n\in\mathbb{N}$. In particular, each $X_n$ is embedded in $X$ as a closed subspace.
Since any compact metric space is separable, the following will certainly apply to an expanding sequence of compact metric spaces.
Theorem The colimit $X$ of an expanding sequence $X_1\subseteq X_2\subseteq\dots$ of closed embeddings of separable metric spaces is a perfectly normal, paracompact $T_2$ space. Moreover, the following three statements are equivalent.
- $X$ is metrisable.
- $X$ is second-countable.
- $X$ is first-countable.
- Each point of $X$ has a neighbourhood contained in some $X_n$.
If each $X_n$ is locally compact, then the above are furthermore equivalent to
- $X$ is locally compact.
The proof will run as follows. First we establish that $X$ is perfectly normal $T_1$, where a space is perfectly normal if it each of its closed subsets is the zero-set of a real-valued function. Since perfect normality implies normality, it will follow that $X$ is regular Hausdorff.
Proposition Every regular Lindelöf space is paracompact. $\qquad\blacksquare$
Thus the paracompactness of $X$ will follow if we can show that it has the Lindelöf property. Recall that a space is Lindelöf if each of its open covers has a countable subcover. Note that a space which is both perfectly normal and paracompact is hereditarily paracompact.
At this stage we have explained how to prove the first part of the proposition. It remains only to explain the equivalence of the claimed metrisation conditions.
$1\Rightarrow2$ follows from the fact that every Lindelöf metric space is second-countable, and the converse is found in the famous Urysohn Metrisation Theorem.
Theorem (Urysohn Metrisation Theorem) A second-countable Hausdorff space is metrisable if and only if it is regular. $\qquad\blacksquare$
Clearly every second-countable space is first-countable. Thus to complete the proof from here, it is enough to show that the first-countability of $X$ implies condition $4$, and that condition $4$ implies the metrisability of $X$. The first part of this is contained in the following lemma.
Lemma Let $X$ be the colimit of a sequence of closed embeddings of $T_1$ space $X_1\subseteq X_2\subseteq\dots$. The following statements hold.
- If $K\subseteq X$ is compact, then $K$ is contained in some $X_n$.
- If $Y$ is a first-countable space and $f:Y\rightarrow X$ a continuous map, then each $y\in Y$ has a neighbourhood $V\subseteq Y$ such that $f(V)$ is contained in some $X_n$.
Proof (1) If $K$ is finite, then the statement is obvious, so assume it is infinite and not contained in any $X_n$. We can assume that the $X_n$ are all distinct and that $K\cap X_1\neq\emptyset$. For each $n\geq1$ choose $x_n\in K\cap(X_n\setminus X_{n-1})$, where we understand $X_{0}=\emptyset$. Write $D=\{x_n\mid n\in\mathbb{N}\}\subseteq X$. If $C\subseteq D$ is any subset, then $C$ is closed in $D$, since for each $n$ we have that $C\cap X_n$ is finite and hence closed in the $T_1$ space $X_n$. It follows that $D$ is infinite, discrete and closed in $X$, and this contradicts the compactness of $K$.
(2) Let $\{V_n\subseteq Y\}_\mathbb{N}$ be a neighbourhood basis at $y\in Y$ such that $V_n\subseteq V_{n-1}$. Assume that $f(V_n)$ is not contained in any $X_n$, and for each $n>1$ choose a point $y_n\in V_n\setminus f^{-1}(X_n)$. This yields a compact subset $D=\{y_n\}_\mathbb{N}\subseteq Y$. By part $(1)$ of the lemma, $f(D)$ is contained in some $X_m$, which implies that $f(y_m)\in X_m$. But this is a contradiction, so it must be that $f(V_n)\subseteq X_n$ for some $n$. $\qquad\blacksquare$
As explained above, at this stage we need only demonstrate that $4\Rightarrow1$. For this we need the following.
Proposition A paracompact Hausdorff space which is locally metrisable is metrisable. $\qquad\blacksquare$
Proofs for all results quoted above can be found in [1]. I'll be happy to dig out page numbers if it's useful to you.
Proof of the proposition That $X$ is $T_1$ is a triviality given that it carries the colimit topology: the points of $X$ are closed because they are closed in each $X_n$.
We prove that $X$ is perfectly normal. Thus let $A\subseteq X$ be a closed subset. We must construct a map $f:X\rightarrow [0,1]$ such that $f^{-1}(0)=A$. To begin, use the fact that the metric space $X_1$ is perfectly normal to find $f_1:X_1\rightarrow [0,1]$ with $f^{-1}_1(0)=A\cap X_1$ (the function $f_1(x)=d(x,A\cap X_1)$ will do, where $d$ is some metric on $X_1$). Now suppose that $f_n:X_n\rightarrow [0,1]$ has been constructed for some $n\geq1$ such that $f_n^{-1}(0)=A\cap X_n$.
Then $A\cap X_n$ is closed in both $X_n$ and $A\cap X_{n+1}$. Consequently, the function $f'_{n+1}:X_n\cup(A\cap X_{n+1})\rightarrow[0,1]$ which is $f_n$ on $X_n$ and $0$ on $A\cap X_{n+1}$ is well-defined and continuous. Since $X_n\cup(A\cap X_{n+1})$ is closed in the metric space $X_{n+1}$, this function has a continuous extension to $f_{n+1}:X_{n+1}\rightarrow [0,1]$ which is positive on the complement of $X_n\cup(A\cap X_{n+1})$.
Thus by induction we construct $f_n:X_n\rightarrow [0,1]$ for all $n\in\mathbb{N}$ such that $f_n^{-1}(0)=A\cap X_n$ and $f_{n+1}|_{X_n}=f_n$. Let $f:X\rightarrow[0,1]$ be the map supplied by the universal property of the colimit. Then $f^{-1}(0)=A$. We conclude that $X$ is perfectly normal.
We show now that $X$ is Lindelöf. Thus let $\mathcal{U}$ be an open cover of $X$. For each $n\in\mathbb{N}$, $X_n$ is a closed subspace of $X$. Since $X_n$ is a separable metric space, it is Lindelöf. Therefore there is a countable subfamily $\mathcal{U}_n\subseteq\mathcal{U}$ which covers $X_n$.
Put $\mathcal{V}=\bigcup_{n\in\mathbb{N}}\mathcal{U}_n$. Clearly $\mathcal{V}$ is a subfamily of $\mathcal{U}$ which covers $X$. Since a countable union of countable sets is itself countable, it follows that $\mathcal{V}$ is countable. We may conclude that $X$ is Lindelöf.
Now we are in a position to address the equivalence of the first three listed properties. Clearly
$$\text{metrisable Lindelöf} \Leftrightarrow \text{second-countable regular} \Rightarrow \text{first-countable}$$
On the other hand, suppose that $X$ is first-countable. Then by the lemma above, any $x\in X$ has a neighbourhood $U\subseteq X$ contained in $X_n$ for some $n$. This is exactly condition $4$.
So, assuming condition $4$, we have for any $x\in X$ an open subset $U\subseteq X$ such that $x\in U\subseteq X_n$ for some $n\in\mathbb{N}$. But $X_n$ is a metric space, so $U$ is metrisable. Thus it follows that each point of $X$ has a metrisable neighbourhood. i.e. that $X$ is locally metrisable. But a paracompact space which is locally metrisable is metrisable, so we get the implication $4\Rightarrow1$ for $X$.
Finally, assume that each $X_n$ is locally compact. If $X$ satisfies condition $4$, then any given $x\in X$ has an open neighbourhood $U$ which is contained in some $X_n$. Since $X_n$ is locally compact Hausdorff and $U$ is its open subset, $U$ is locally compact. Thus $x$ has a compact neighbourhood $K$ in $U$. But $U$ is open in $X$, so $K$ is also a compact neighbourhood of $x$ in $X$. This shows that each point of $X$ has a compact neighbourhood. Since $X$ is Hausdorff, it is thus locally compact.
On the other hand, if $X$ is locally compact, then any $x\in X$ has a compact neighbourhood $K$. Since $K$ contains in some $X_n$ by the lemma proved above, condition $4$ is clearly satisfied. $\qquad\blacksquare$
Example: Let
$$X_1\subseteq X_1\subseteq X_2\subseteq\dots$$
be an expanding sequence of inclusions of closed submanifolds of positive codimension.
Claim The colimit $X$ of the sequence is not metrisable. In particular, it is not first-countable, second-countable, or locally compact.
For suppose that $x\in X$ has a neighbourhood $U$ contained in some $X_n$. We can assume that $U$ is a manifold chart of $x$ in $X_n$. Since $X_n$ is a submanifold of $X_{n+1}$ we can furthermore assume that this is the restriction of an adapted chart of $x$ in $X_{n+1}$. Now, $U$ is open in $X$, so $U$ is also open in $X_{n+1}$. But contradicts invariance of domain, since $X_n$ is a closed submanifold of $X_{n+1}$. In particular, no such neighbourhood $U$ exists. $\qquad\square$
Clearly the content of the example applies to the classifying spaces $BO_k$.
Remark Any expanding sequence of closed embeddings of metric spaces is paracompact. Separability is not required for this, but the argument is slightly more involved. In such a situation, the conditions $1,3,4$ (and $5$) remain equivalent.
References
- R. Engelking, General Topology, second edition, Heldermann Verlag Berlin, (1989).
