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Learning about dense sets the classical example is that of $\mathbb Q$, the rationals, in $\mathbb R$. The same interpretation is valid for irrationals in $\mathbb R$. I was wondering if a dense set needs to be infinite, because this is what intuition would suggest. Moreover, are dense sets always countably infinite?

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  • $\begingroup$ $\mathbb R$ is a dense subset of itself. The irrationals are a dense subset of $\mathbb R$. $\endgroup$ – lulu Feb 3 at 15:57
  • $\begingroup$ Even if we restrict to, lets say, the interval $[0,1]$ , no finite set can be dense in that interval. $\endgroup$ – Peter Feb 3 at 16:03
  • $\begingroup$ The definition of a topological space is very broad, that is, it is not very restrictive. In the discrete topology on a set S, every subset of S is open (and also closed), and the only dense set is S..... On the other hand if S is a sub-space of $\Bbb R$ (with the usual topology), then S does have at least one countable dense subset. $\endgroup$ – DanielWainfleet Feb 4 at 2:13
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To answer the general question of "are dense sets always infinite": no, because certainly, if $X$ is a finite topological space, then $X$ is dense in itself.

For another example, if $X$ is any set with the indiscrete topology, then every nonempty subset of $X$ is dense.

For yet another example, let $X = \mathbb{R}$ with the topology determined by the Kuratowski closure operator $$\operatorname{cl}(S) = \begin{cases} S, & 0 \notin S; \\ \mathbb{R}, & 0 \in S.\end{cases}$$ Then $\{ 0 \}$ is dense in $X$, yet $X$ is $T_0$. (In fact, this example can easily be modified to give a $T_0$ topological space of any desired cardinality such that some single point is dense.)

On the other hand, in a $T_1$ topological space, every finite subset is closed. So, if a finite subset of a $T_1$ topological space $X$ is dense, then $X$ itself must be finite. Or, for the contrapositive, if $X$ is an infinite $T_1$ topological space, then every dense subset of $X$ is infinite.

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    $\begingroup$ Your "certainly" is dreadfully ambiguous here, and I imagine some non-native speakers will be scratching there heads over it. One meaning of "certainly" is "yes, definitely"! I suggest replacing it with "no" :-) $\endgroup$ – TonyK Feb 3 at 22:33
  • $\begingroup$ Re: "if a finite subset of a $T_1$ topological space $X$ is dense, then $X$ itself must be finite": More specifically, $X$ itself must be that finite subset. A finite proper subset of a $T_1$ space cannot be dense. $\endgroup$ – ruakh Feb 3 at 22:58
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    $\begingroup$ I took the liberty of changing "certainly" to "no, because certainly". $\endgroup$ – Tanner Swett Feb 4 at 1:00
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    $\begingroup$ Ryan's answer illustrates my point beautifully! $\endgroup$ – TonyK Feb 4 at 10:19
  • $\begingroup$ @TonyK Good point, agreed. $\endgroup$ – Daniel Schepler Feb 4 at 18:56
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A dense set of the reals is always infinite. It need not be countable: For example, the reals are dense in themselves.

And a dense set needs not be infinite either. For example, $\{1\}$ is a finite set that is dense in itself.

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  • $\begingroup$ This is true for the reals with the usual topology, but it isn't true for other topologies (like the indiscrete topology mentioned in the accepted answer). $\endgroup$ – probably_someone Feb 4 at 2:22
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Certainly a set dense in the real line must be infinite: if $X$ is a finite subset of the real line, it has some maximum element $m$. Then no point in $X$ is close to $m+1$ (otherwise $m$ is not the maximum, contradiction). So $X$ cannot be dense.

On the other hand, you mention that the irrational numbers are also dense, but these are already uncountable, so there’s no reason to think a dense set should be countable. (In fact, the more important thing is whether there is a countable dense set. This is called separability; see https://en.m.wikipedia.org/wiki/Separable_space).

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  • $\begingroup$ Did you answer the question "Is a dense set always infinite?" It sounds like your first paragraph is saying "Certainly yes," but you restrict your explanation to the real line, with no indication of whether you think the answer applies to other situations. $\endgroup$ – LarsH Feb 4 at 12:56
  • $\begingroup$ You’re right, but there are better answers now so $\endgroup$ – Ryan Feb 4 at 15:59
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Every dense subset of $\mathbb R$ is infinite. A finite set has a maximum so it clearly cannot be dense. Dense subsets of $\mathbb R$ need not be countable; for example, the irrational numbers and the entire set $\mathbb R$ are both uncountable and dense.

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In an infinite metric space any dense set must be infinite too: all finite sets are closed and so their closure ( the same finite set) cannot equal the whole space. One dense set is the whole space, so it need not be countable, unless the whole space is.

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