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I am trying to show that $\mathbb{R}^\infty$ with direct limit topology is not metrizable by showing that it is not paracompact via the following theorem:

Thm: A locally metrizable space $X$ is metrizable iff $X$ is Hausdorff and paracompact.

Clearly $\mathbb{R}^\infty$ is locally metrizable (take any sufficiently large $n$ and consider the neighborhood $\mathbb{R}^n$ around a point) and Hausdorff. To show that $\mathbb{R}^\infty$ is not paracompact, I took the cover $\{ \mathbb{R}, \mathbb{R}^2, \ldots\}$ and tried to show that there is not an open refinement such that it is locally finite. I figured this is the most natural open cover to work with, but it is hard to construct such an open refinement. Of course this doesn't have to be the only open cover we have to consider if we are trying to disprove paracompactness.

Any help is appreciated!

green frog
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    In fact, $\mathbb{R}^\infty$ is paracompact. – Eric Wofsey Feb 14 '22 at 04:58
  • @EricWofsey, in general is the direct limit of paracompact spaces, paracompact? – green frog Feb 14 '22 at 05:09
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    Not in full generality. I believe it is true that the direct limit of a sequence of closed inclusions of paracompact Hausdorff spaces is paracompact, though I don't recall the full details of a proof off the top of my head. (The idea would be to construct a partition of unity subordinate to a given open cover inductively over each space in the sequence.) – Eric Wofsey Feb 14 '22 at 05:29
  • If $X$ is Tychonoff, then $X$ is paracompact iff $X\times K$ is normal for each compact Hausdorff $K$. This is useful eg. because $(-)\times K$ commutes with colimits. – Tyrone Feb 14 '22 at 08:43

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Actually, $\mathbb{R}^\infty$ is not locally metrizable; I'll try to explain why your argument doesn't work.

If we want to prove that a space $X$ is locally metrizable, we need to show that every point of $X$ has a metrizable neighborhood. This means that the first step in our proof must be to take an arbitrary point $x = (x_i)_{i \in \mathbb{N}} \in \mathbb{R}^\infty$. Then we have some $n \in \mathbb{N}$ such that $x_i = 0$ for all $i > n$.

Perhaps this is (so far) exactly what you meant, but I think it's important to have it be clear through your explanation that the value of $n$ depends on the point $x$ (imo, this was not clear in your explanation).

Anyway, next you wanted to consider the neighborhood $\mathbb{R}^n$ around the point $x$. This doesn't work – in fact there are no neighborhoods of any point whatsoever in $\mathbb{R}^\infty$ which are homeomorphic to $\mathbb{R}^n$. (Intuitively, this should make sense: there also aren't any neighborhoods of any point in $\mathbb{R}^2$ which are homeomorphic to $\mathbb{R}^1$). If I had to guess, you are implicitly claiming here that $U := \{(y_i)_{i \in \mathbb{N}} \in \mathbb{R}^\infty : y_i = 0 \; \forall i > n\}$ is a neighborhood of $x$, but it is not. Try to work out why by using the definition of the direct limit topology!

diracdeltafunk
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  • Very helpful, thank you. Your definition of $U$ was exactly what I had in mind. We can take any open ballof 0 in $\mathbb{R}^{n + 1}$ and see that any points with nonzero $n + 1$-th coordinate won't be contained in $U$, hence $U$ cannot be open in the direct limit topology. Is this a valid argument? – green frog Feb 14 '22 at 05:02
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    Yeah that's the right idea! Just to make sure it's clear for you: when you say that "points with nonzero $n+1$-th coordinate won't be contained in $U$", this more precisely means "point with nonzero $n+1$-th coordinate won't be contained in $i^{-1}(U)$", where $i = (x_1, \dots, x_{n+1}) \mapsto (x_1, \dots, x_{n+1}, 0, 0, \dots) : \mathbb{R}^{n+1} \to \mathbb{R}^\infty$ is the canonical inclusion. – diracdeltafunk Feb 14 '22 at 06:13