Maybe someone familiar with Willard's textbook can help me out. Problem section 23G on pg. 174 is titled piecewise metrizability. The first problem is:

  1. If a Tychonoff space $X$ is the union of a locally finite collection of closed, metrizable subspaces, then X is metrizable.

This fact is well-known. It is 4.4.19 in Engelking's book. It also appears in Nagata's seminal metrization paper from 1950. However, the conclusion is true without the assumption that $X$ is Tychonoff. This leads me to think there is a simpler proof under this assumption. If you know of one, please give me an outline or hint.

More interesting is the next problem from Willard:

  1. If a $T_4$ space $X$ is the union of any locally finite collection of metrizable subspaces, then $X$ is metrizable. [Use 15.10]

This is less well-known, because it is clearly not true. Consider the one point compactification of an uncountable discrete space. It is normal and non-metrizable, but is the union of just two metrizable subspaces: one discrete and the other a singleton. The hint refers to the standard "shrinking" theorem for point-countable open covers of normal spaces. So, it seems likely that the word "open" was just accidentally omitted. The following fact follows easily from the first exercise and 15.10:

Fact: If a $T_4$ space $X$ is the union of a locally finite collection of open, metrizable subspaces, then $X$ is metrizable.

Certainly, someone must have noticed this before now. However, I haven't been able to find any reference to, or use of, the above fact in print. Does anyone know of a better citation for this than "Corrected version of exercise 2 in 23G of [W]"?

Thanks in advance,


  • Willard, Stephen General topology Originally printed 1970 by AddisonWesley; currently in print by Dover Publications (2004)
  • Engelking, Ryszard General topology Heldermann Verlag (1989)
  • Nagata, Jun-iti On a necessary and sufficient condition of metrizability J. Inst. Polytech. Osaka City Univ. (1950)

PS: Unfortunately, exercises 1 and 2 also appear (without correction) in Patty's more recent textbook, Foundations of Topology, Jones and Bartlett Publishers (2009). They appear to have been copied verbatim from Willard.


2 Answers 2


Your corrected version of exercise 2:

Suppose $X$ is $T_4$ and $X=\bigcup \mathcal{U}$ where all $U \in \mathcal{U}$ are open and metrisable, and $\mathcal{U}$ is locally finite. Then $X$ is metrisable.

This follows right away from the standard fact that we can shrink the open cover to a closed one (in a $T_4$ space we can shrink all point-finite covers and a shrinking of $\mathcal{U}$ is still locally finite) and apply part 1. It seems quite a logical progression from 1 to me.

I don't as yet see how Willard envisioned the proof of the closed case in the extra presence of complete regularity; it seems that we would hint towards a uniform spaces based metrisation theorem, as Nagata did (so Willard says in the accompanying text)?

  • $\begingroup$ Right, sorry, I should have been more clear. Willard's 15.10 is the shrinking theorem you refer to, which clearly makes the corrected exercise an easy one. I'm still guessing that this fact probably appear somewhere. The "metrization industry" has produced a vast literature in the past half-century or so. $\endgroup$ Apr 12, 2019 at 1:27
  • $\begingroup$ Also, shortly after I commented on the first post, I realized another solution to the "corrected exercise." While this doesn't hold in general, I believe that if $X$ is a normal space, then the union of any locally-finite collection of open paracompact subspaces of $X$ is paracompact. So the $X$ in the exercise is metrizable because it is paracompact and locally metrizable. $\endgroup$ Apr 12, 2019 at 1:33
  • $\begingroup$ @JeffNorden do you have a reference for the $T_4$ open case? Any idea how to do 1 within Willard’s framework ? Engelking reduces it to the closed map theorem, but IIRC Willard doesn’t cover that. $\endgroup$ Apr 12, 2019 at 16:08

This is more of a comment (which I don't seem to be allowed to post):

Since the corrected version of (2) is an immediate (even trivial) corollary of the Nagata–Smirnov metrization theorem, I would wager, if it does appear somewhere, it occurs as an aside or footnote.

That said, the corrected statement of (2), vaguely resembles the forward direction of the Smirnov metrization theorem (i.e. paracompact Hausdorff and locally metrizable, iff, metrizable)

  • $\begingroup$ Yes, it is tempting to prove the above Fact from Nagata-Smirnov, but... For one thing, the normality of $X$ is required. There is a non-normal space that is the union of just two open metrizable subspaces. This was discovered, not surprisingly, by R.H. Bing. See the next comment for more details. $\endgroup$ Apr 9, 2019 at 15:17
  • $\begingroup$ The problem in the proof you were probably thinking of is the following. If $\cal U$ is a locally finite open cover of $X$ and you then have a relatively-locally-finite open cover of each $U\in\cal U$ (i.e., locally finite in the subspace $U$ of $X$), it does not follow that the union of these covers is locally finite in $X$. In other words, the union of a locally-finite collection of open paracompact subspaces can fail to be paracompact. See the next comment as well. $\endgroup$ Apr 9, 2019 at 15:34
  • 1
    $\begingroup$ The problem occurs at a point $x$ in the boundary of some $U$. The paracompactness of $U$ doesn't "know" about $x$, so an open susbset of $X$ that contains $x$ can intersect as many members of a locally-finite cover of $U$ as it wants to. On the other hand, this isn't a problem for point-finite covers. The union of a point-finite collection of open pointwise-paracmpact (aka metacompact) subspaces will be pointwise-paracompact. $\endgroup$ Apr 9, 2019 at 15:34

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