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Under what circumstances is the discrete metric space separable? Can anyone help me please?

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    Does a countable discrete space contain a countable dense set? Does an uncountable discrete space contain a countable dense set?2012-06-13
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    As I've emphasized to you numerous times, Miguel, you **need** to explain what you've tried so far. People are far less inclined to help those who have not put in any effort themselves, so demonstrate that you have by explaining what you've thought about regarding this problem so far.2012-06-13
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    I think 2 down votes were enough. Keeping on piling is not going to express your sentiment any clearer to the OP.2012-06-13
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    I think this is an example to R. I'm not complaining, just answer the questions they ask2012-06-13

3 Answers 3

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Recall that:

  1. If $A$ is a dense and closed subset of $X$ then $A=X$.
  2. In a discrete space every set is open, therefore every set is closed.
  3. If $X$ is discrete and separable then there is a countable subset which is dense.
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Hint: Let $M$ be a metric space with the discrete metric.

  • When is a subset $S\subseteq M$ closed?

  • What is the closure of a set $S\subseteq M$?

  • Thus, which subsets of $M$ are dense?

  • When is there a countable dense subset of $M$?

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A space $X$ is separable if it contains a dense countable subset $D$.

Now that we know the definition we need to think about what it means for $D$ to be dense in a discrete space. Dense means that if we pick any point $x$ in $X$ and an open set $O$ containing it, then $O$ will intersect with $D$.

In a discrete space, the singleton set $\{x\}$ is open. The only way this set can have non-empty intersection with $D$ is if we have $x \in D$.

But this means that the only dense subspace of a discrete space $X$ is $X$ itself. Hence, the only way to have a countable dense subset of a discrete space is if the space itself is countable.