TITLE: Countable and Uncountable Sets in Metric Spaces
QUESTION [0 upvotes]: I've been thinking about metric spaces recently, and a very abstract question came to mind. Please bear with me as I try to articulate it.
In a metric space, if there exists an uncountable set of vectors, can I choose a countable set of vectors in the space such that no element of my set is an element of the uncountable set? 
As a follow up, I have questions about my question...

Am I even allowed to ask this question
Is this a question that can be answered in general? Are there certain conditions under which the answer would be "yes"?

REPLY [2 votes]: Your question is a little bit ambiguous depending on what you mean by 'choose'.
So first I'll answer this: "If there exists an uncountable set of vectors $A$, is there a non-empty countable set of vectors in the space such that no element of my set is an element of $A$?"
Clearly, the answer is no. For let the set $A$ be all the vectors of the metric space $X$. Then there is no non-empty countable set of vectors of $X$ that is not included in $A$.
The other question, which I guess is the one you were trying to ask is: "If there exists an uncountable set of vectors $A$, and there is a non-empty countable set of vectors in the space such that no element of my set is an element of $A$, how could I define ('choose') such a set?"
The answer is as follows. Let $X$ be the set of all the vectors, since $A$ is not $X$, then $X-A$ is non-empty. Now pick a $x \in X-A$. The set $\{x\}$ alone is a countable set not included in $A$, but you probably meant a countably infinite set. If $X-A$ is finite, then obviously you can't define a countably infinite set not in $A$. If $X-A$ is infinite you can do a construct a set $B$ that is countably infinite as follows:

If $X-A$ is unbounded, then pick any $x_0 \in X-A$ and keep adding members from bigger and bigger balls around $x_0$ (e.g., balls of radius $1/n$). As long as you enforce the new members being different (if possible) from the ones that are already inside it will work.
If $X-A$ is bounded, then pick an accumulation point $x_0$ of $X-A$, and keep adding balls from smaller and smaller balls around $x_0$ (e.g., balls of radius $1/n$). The same logic as above applies.

You are done. This became bigger than I expected but I hope it's clear :)
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