metric space $X$ in which a finite set is dense

Question

My question is,what can you say about a metric space $X$ in which a finite set is dense?

My thought: Let $A$ be any finite set in $X$.If $X$ is infinte then for any $x$ in $X$ and for any $p>0$, $d(a,x)<p$ for some $a$ in $A$.Since $A$ is finite,we must have,$x$ is in $A$.This implies that $x$ belongs to $A$ for all $x$ in $X$,a contradiction.Hence $X$ cannot be infinte.

Is my process correct?

What process? You didn't even draw a conclusion. – Jan 10 '18 at 17:31 — , Jan 10 '18 at 17:31
My conclusion is that X must be finite. – jimm Jan 10 '18 at 17:36 — jimm, Jan 10 '18 at 17:36

score 2 · Accepted Answer · answered Jan 10 '18 at 17:36

2

Yes, your argument is Ok.

As an alternative, assuming you've proved that singletons in a metric space are closed, it follows that any finite set is closed. Hence, if a finite set is dense, it must be the full space.

answered Jan 10 '18 at 17:36

quasi

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score 1 · Answer 2 · answered Jan 10 '18 at 17:36

1

Let $X$ be a metric space. Then, any finite set is automatically closed. So, if $A$ is a dense, finite subset, we have $A = \overline{A} = X$.

answered Jan 10 '18 at 17:36

Duncan Ramage

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metric space $X$ in which a finite set is dense

2 Answers2