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Let $C$ bethe Cantor set, then is it true that $C = \text{bd}(C)$?

I know that the $C$ is closed since it the intersection of closed intervals, which is always closed. This means that $C$ contains $\text{bd}(C)$. To show that $C = \text{bd}(C)$ I would need to show that $\text{bd}(C)$ contains $C$. How should I do this?

Asaf Karagila
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fosho
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    Use the fact that the boundary is the closure minus the interior. Now, since $C$ is closed we know that it is its own closure, so it now suffices to show that the interior is empty... – Tommy Tang Mar 22 '16 at 10:22

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Let $P\in\mathcal C$ and let $B$ a ball centred in $P$. Since the cantor set is perfect, if $x\in B$ is an element of the cantor set, then, there is $y\in ]x,P[$ s.t. $y$ is not in the cantor set. Therefore, the interior of the cantor set is empty. This prove the claim.

Surb
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