In a textbook, the following is written.

I have a complete metric space, $S$. I have a union of a sequence of nowhere dense sets. If each set in the sequence is replaced by its closure, then the union will ‘only get larger’.

This appears strange, given that the closure of each of these sets will have empty interior (seeing as the sets are nowhere dense).

I am having some trouble understanding why this is so. Any clarification is much appreciated.

  • $\begingroup$ The closure of a set contains the set itself, as $\{s,s,s\cdots\}$ is a sequence which converges to $s$ (if you prefer, some definitions explicitly include the set into its closure, as here) $\endgroup$
    – lulu
    Aug 19, 2021 at 17:46
  • $\begingroup$ The closure of a set is the union of the set and the set’s boundary. I.e. a set is a subset of it’s closure… $\endgroup$ Aug 19, 2021 at 18:03
  • $\begingroup$ I think you are misreading your textbook. Please give a reference or a more detailed quotation. The claim as stated is true but rather odd. $\endgroup$
    – Rob Arthan
    Aug 19, 2021 at 20:35
  • $\begingroup$ Maybe you are confused by the notion that $A$ is dense in $\overline{A}$, yet $A$ was nowhere dense? Different topologies. $A$ is nowhere dense in the topology of $S$, but dense in the subspace topology on $\overline{A}$ induced by the topology of $S$. $\endgroup$ Aug 20, 2021 at 5:29
  • $\begingroup$ First, you say "I have a union of a sequence of nowhere dense sets." Then, you say "(seeing as the sets are dense)." Which is it? Are the sets dense or nowhere dense? $\endgroup$
    – 5xum
    Aug 20, 2021 at 5:39

1 Answer 1


The closure $\overline A$ of a set $A\subseteq S$ is, almost by definition, at least as large as $A$ itself. There are several definitions of closure, but here are three common (equivalent) definitions:

The closure $\overline A$ of $A$ is the union of $A$ with its boundary.

The closure $\overline A$ of $A$ is the smallest closed set that contains $A$.

The closure $\overline A$ of $A$ is the intersection of all closed sets that contain $A$.

The first two are obviously larger than $A$, and the third is only slightly less obvious.

So if we have a sequence of sets $A_1, A_2, A_3, \ldots$, then $A_1\subseteq \overline{A_1}, A_2\subseteq \overline{A_2}, A_3\subseteq \overline{A_1}$, and so on. And this implies $A_1\cup A_2\cup A_3\cdots\subseteq \overline{A_1}\cup\overline{A_2}\cup\overline{A_3}\cdots$.

The fact that the sets are nowhere dense is completely irrelevant to this particular argument.

  • $\begingroup$ The question as asked is more bizarre than the question that I think you are answering: each dense set in the sequence is to be replaced by its closure, but the sets are all nowhere dense, so no set is changed by these replacements. $\endgroup$
    – Rob Arthan
    Aug 19, 2021 at 20:39
  • 1
    $\begingroup$ @RobArthan Nice catch. I didn't see that at all. Yeah, that seems like a strange thing to write in a book. Or a misquote. Either way. $\endgroup$
    – Arthur
    Aug 19, 2021 at 20:58
  • $\begingroup$ Apologies, I did not mean to write that each dense set is to be replaced by its closure, but rather that each set (I.e. each nowhere dense set) in the sequence, is. $\endgroup$
    – Charles
    Aug 20, 2021 at 5:04

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