Let $(X,d)$ be a M.S. without any isolated points and $A$ be a subset of $X$ such that each point is an isolated point of it. Show that $A$ is Nowhere dense.
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A need not be closed. Consider ${1/n\mid n\in\Bbb N}$ – Stefan Hamcke Nov 02 '13 at 00:32
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Yes, you're right! – UNM Nov 02 '13 at 00:37
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If $\overline{A}$ has nonempty interior, then there is $x\in X$ and $r>0$ s.t. $B(x,r)\subset \overline{A}$. In general $x$ is not isolated, but we can assume that $x$ is isolated because if $x$ a limit point then we can take some $y\in B(x,r)\cap A$ and $y$ is isolated, and $y\in B(x,r)$ so we can take $r_0>0$ s.t. $B(y,r_0)\subset B(x,r)$. Since $x$ is isolated, we get $B(x,r)\cap \overline{A}=\{x\}$. Since $B(x,r)\subset \overline{A}$, we get $B(x,r)=\{x\}$, and therefore $\{x\}$ is open in $X$. However if $X$ does not have isolated point then every singleton is not open.
Hanul Jeon
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To prove that $A$ is nowhere dense: suppose it is not, and pick $x\in\mathrm{int}(\overline{A})$. Then, there exists an $r>0$ such that the ball $B(x,r)$ is a subset of $\overline{A}$. But, $x\in\overline{A}$, so $A\cap B(x,r)\neq\emptyset$, so there exists $y\in B(x,r)\cap A$. $B(x,r)$ is open, so there exists $s>0$ such that $B(y,s)\subseteq B(x,r)$: then, $y\in A$, and $B(y,s)\subseteq\overline{A}$.
Now, $y\in A$, and $A$ has no isolated points, so there exists $s_1<s$ such that $B(y,s_1)\cap (A\setminus\{y\})=\emptyset$, so there are no other points of $\overline{A}$ in $B(y,s_1)$, except for $y$. But $B(y,s_1)\subseteq B(y,s)\subseteq\overline{A}$, so this shows that $B(y,s_1)=\{y\}$. But this is a contradiction, since $X$ has no isolated points.
detnvvp
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