Off the top of my head, I can't think of a non-separable compact space.
Can you provide a good example?
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- 2$\begingroup$ That's because any compact metric space is separable. So any example has to be non metrizable $\endgroup$Giuseppe Negro– Giuseppe Negro2014-05-07 15:03:50 +00:00Commented May 7, 2014 at 15:03
- 3$\begingroup$ Check dantopology.wordpress.com/2012/10/13/alexandroff-double-circle $\endgroup$Ugo Iaba– Ugo Iaba2014-05-07 15:07:15 +00:00Commented May 7, 2014 at 15:07
- $\begingroup$ Thanks Léo & Giuseppe. The Alexandroff double circle is non-separable and compact. $\endgroup$Tom LaGatta– Tom LaGatta2014-05-07 15:25:59 +00:00Commented May 7, 2014 at 15:25
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5 Answers
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2 $\pi$-Base, an online version of the reference chart from Steen and Seebach's Counterexamples in Topology, gives the following examples of compact, non-separable spaces. You can view the search result for more details on these spaces.
Alexandroff Square
Closed Ordinal Space $[0, \Omega]$
Concentric Circles
Either-Or Topology
Lexicographic Ordering on the Unit Square
The Extended Long Line
Tychonoff Plank
Uncountable Excluded Point Topology
Uncountable Fort Space
Uncountable Modified Fort Space
- $\begingroup$ Awesome list, Austin! $\endgroup$Tom LaGatta– Tom LaGatta2014-05-08 03:58:11 +00:00Commented May 8, 2014 at 3:58
- $\begingroup$ The long line is my favourite example of this phenomenon! $\endgroup$TiddSchmod– TiddSchmod2018-12-15 03:07:00 +00:00Commented Dec 15, 2018 at 3:07
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A not so simple, but curious example is the Alexandroff Double Circle:
Consider $C_i=\{(x,y) \in \mathbb{R}^2; x^2+y^2=i\}$ for $i=1,2$. Let $X=C_1 \cup C_2$ and $f:C_1 \to C_2$ the radial homeomorphism. Define a topology on $X$ as follows: the points in $C_2$ are all isolated; for each $x \in C_1$ and $n \in \mathbb{N}$ let $O(x,n)$ the arc in $C_1$ centered at $x$ and with length $\frac{1}{n}$. Now take $B(x,n)=O(x,n) \cup f(O(x,n)-\{x\})$ as open neighborhood. The picture below illustrate such $B(x,n)$.
This way we form a basis for a topology on $X$. $X$ provided with this topology is what we call the Alexandroff Double Circle.
Now, $X$ is not separable because all points in $C_2$ are isolated, hence no countable subset of $C_2$ is dense in $C_2$. For compactness, take an open cover $\{A_{\alpha}\}$ of $X$ which consists of basis neighborhoods. Then it is also an open cover of $C_1$. But notice that the topology of $C_1$ as a subspace of $X$ is exactly the usual one. So it is compact and we can find a finite subcollection $\{A_1,\dots,A_k\} \subset \{A_\alpha\}$ that covers $C_1$. Now, by the nature of our neighborhhods, $X-(A_1\cup \dots \cup A_k)$ is a finite set, so we take an extra $A_{\alpha}$ for each of these points, obtaining a finite subcover.
I don't know if this space is useful for anything other than counterexamples. This and a lot of other crazy spaces can be found on "Counterexamples in topology", by L. A. Steen and J. A. Seebach.
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0 This I think would be the simplest example. Let $(Y,\tau_d)$ be an uncountable discrete space. Consider the space $X=Y\cup \{x\}$ with topology $\tau=\tau_d \cup \{X\}$. $(X,\tau)$ is clearly compact and, since all points in $Y\subset X$ are isolated, not separable.
The space I described is $\text{T}_0$ but not $\text{T}_1$. @Léo 's answer gives an example of a compact Hausdorff space that is not separable.
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2 A space $X$ which is product of a collection of compact Hausdorff spaces $\{C_j\}_{j \in J}$ such that the cardinality $|J|$ of the collection is very large. $X$ is compact by Tychonoff's theorem. If $X$ is separable, i.e. if there is a countable dense subset $A$, then every point of $X$ is the limit of a sequence of points in $A$. The number of such sequences is at most the cardinality of the reals, and each sequence has at most one limit because $X$ is Hausdorff, and so $X$ has cardinality at most equal to the cardinality of the reals. But you can make $X$ have larger cardinality by choosing $|J|$ sufficiently large.
Edit: As pointed out in the comment of @Henno Brendsma, I should not have used convergence of sequences since that carries an additional assumption that each point has a countable neighborhood basis. Instead one can just use convergence of nets. The cardinality of all nets in a countable subset $A$ is bounded, so the cardinality of the set of limits of convergent nets in $A$ is bounded, and every point in the closure of $A$ is the limit of a convergent net in $A$.
- 6$\begingroup$ It's not true that in every point of $X$ is the limit of points from a dense subset. This is true for first countable spaces, but large products of $[0,1]$, say, won't be first countable. In fact $[0,1]^I$ is separable iff $|I| \le |\mathbb{R}|$. So your idea does work, but for a different reason. $\endgroup$Henno Brandsma– Henno Brandsma2014-05-07 16:13:47 +00:00Commented May 7, 2014 at 16:13
- $\begingroup$ The cardinality of all subnets of a countable set is not bounded. You need to use filters to get the upper bound of $2^{2^{\aleph_0}}$ for the cardinality of a separable space (because a filter on $X$ is an element of $\mathcal{P}(\mathcal{P}(X))$). This is a known area of deficiency of Moore-Smith convergence. $\endgroup$Robert Furber– Robert Furber2019-11-11 10:20:09 +00:00Commented Nov 11, 2019 at 10:20
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Based on this question I asked recently, the one-point compactification of a non-separable metric space.
I specifically asked about the space of cadlag real-valued functions on $[0,1]$ under the supremum metric, but there are lots of other non-separable metric spaces out there, and other compactifications where the same argument applies.
The metric space is an open subset of its one-point compactification, and is not separable, so the one-point compactification is also not separable.
