Separable space and countable

Question

How do I show that every collection of disjoint open sets in a separable space must be countable?

I am studying separable from this book and this was stated, and the prove was left to the reader. I am still trying to understand the material, so I have some problem with this, can anyone give me a light, please?

score 3 · Answer 1 · edited Jun 12 '20 at 10:38

This may be helpful for your question.

Let us recall that a space $X$ is said to be CCC, i.e., countable chain condition if $X$ satisfies the condition that every collection of disjoint open sets in a space must be countable. Note that

Every separable space is CCC. The converse is not true.

For example:

Let $X$ be a space with $|X|= \aleph_1$, let $\tau_X= \lbrace U: X\setminus U \text{ is countable } \rbrace$. This space is CCC, but not separable.

Proof: $X$ is not separable: for any countable set $A \subset X$, clearly, $U=X\setminus A$ is open and $U \cap A=\emptyset$.

$X$ is CCC: if $X$ has uncountable disjoint open sets $\lbrace \cal U_\xi: \xi \in \aleph_1\rbrace$. Pick one open set, for example, $U_0$. Because $X \setminus U_0$ is uncountable, it is a contradiction with $U_0$ is open.

score 2 · Answer 2 · answered Apr 09 '13 at 20:37

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Can you find an injection from the collection $\{U_i\mid i\in I\}$ of disjoint open sets to the countable dense subset $D$? Then the cardinality of the collection cannot be larger than the cardinality of $D$.

answered Apr 09 '13 at 20:37

Stefan Hamcke

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score 2 · Accepted Answer · answered Apr 09 '13 at 20:40

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Since your space is separable there is a countable subset S which is dense. If your collection of open subsets is uncountable you could choose an element of S from each of these which would contradict the fact that S is countable.

answered Apr 09 '13 at 20:40

John Douma

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what is the relation ship of the dense set S and the collection of uncountable open subsets ? – Akaichan Apr 09 '13 at 20:59
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A dense set intersects every open set. Since the collection is assumed to be disjoint, we can find a unique element of S in each member of the collection. – John Douma Apr 09 '13 at 21:02
and since these sets are assumed to be uncountable, that is a contradiction to the countable set S, right? – Akaichan Apr 09 '13 at 21:06
That is correct. – John Douma Apr 09 '13 at 21:07
Thank you very much, and how does one know that a dense set intersects every open sets, just out of curiosity ? – Akaichan Apr 09 '13 at 21:08
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And you have to assume non-empty open sets so you can pick an element from each intersection which is then non-empty by definition of dense. – Stefan Hamcke Apr 09 '13 at 21:11
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If $S\subset X$ is dense, every point in $X$ is either in $S$ or a limit point of $S$. – John Douma Apr 09 '13 at 21:14

Separable space and countable

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