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If $A$ is everywhere dense in the metric space $M$, then the only closed set which contains $A$ is $M$.

My attempt: Let $B$ be a closed set which contains $A$.

My aim: Prove $B=M$

Since $A$ is everywhere dense in $M$, $\bar{A}=M$. By the theorem 'Closure of a subset $A$ of a metric space $M$ is the intersection of all closed set containing $A$, $M=\bar{A} \subset B$

Regarding the direction $B \subset M=\bar{A}$, I have no idea how to prove a point is in a closure set. Can anyone give some hints?

EDIT: let $b\in B$. Since $B$ is a closed set, for all $\epsilon > 0$, there exists a neighbourhood $N(b,\frac{\epsilon}{2})$ such that it contains a point of $B$ distinct from $b$.

Let $a \in A$. Since $A \subset B$, for all $\epsilon > 0$, there exists a neighbourhood $N(a,\frac{\epsilon}{2})$ such that it contains a point of $B$ distinct from $a$.

Let $x$ be an intersection element between the two neighbourhood. Then $|a-b| \leq |a-x|+|x-b|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$,which implies that $b \in N(a,\epsilon)$, that means $b \in \bar{A}$.

Is my proof correct?

Idonknow
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2 Answers2

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You’re missing the forest for the trees.

You want to show that $B=M$, and you’ve shown that $M\subseteq B$, so you need to show that $B\subseteq M$. But this requires no work at all: $M$ is the space in which all of this is taking place, so every set that you mention is a subset of $M$!

In a little more detail: you know that $B$ is the intersection of all closed subsets of $M$ that contain $A$, and an intersection of subsets of $M$ is certainly a subset of $M$ itself, so $B\subseteq M$. Or, if you prefer, you can observe that $M$ is a closed subset of $M$ containing $A$, so $B$, which is the intersection of all closed subsets of $M$ containing $A$, is in particular a subset of $M$.

Brian M. Scott
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Hint

Use this result: If $A\subset B$ then $\overline{A}\subset\overline{B}$ where $\overline{X}$ is the closure of $X$.