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Is the interior of a connected set in $\mathbb R^k$ connected?

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No, as answers below suggest. But, however you may want to prove that closure of connected sets are connected. (and that, interior of connected sets in $\Bbb{R}$ are connected.) – user21436 May 5 '12 at 20:41
up vote 3 down vote accepted

No. If $X\subset\mathbb R^2$ is the union of two closed disks of radius $1$, one with center at $(1;0)$ and another with center at $(-1;0)$, then $X$ is connected but its interior is not.

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how do you show that the interior is both closed and open? It is eas to see by visualization, but is there a formal way to see this? – user43901 Nov 12 '12 at 0:28
@user43901: Both closed and open in what space? The interior of $X$ is trivially closed and open in itself, and it is not closed in $X$. Neither of those facts have bearing on why the interior is disconnected. It is a union of $2$ disjoint nonempty open subsets. – Jonas Meyer Jul 3 '13 at 6:12

Nope... Pick two tangent balls...

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BTW: probably a better example is two non-touching disks connected by a segment of line. – N. S. May 6 '12 at 1:52

No. Let

$A_1=\{(x,y)|x\leq0,y\leq0\}$ (The third quadrant and the positive x-axis) and

$A_2=\{(x,y)|x\geq0,y\geq0\}$.(The first quadrant and the positive x-axis).

$A_1\cup A_2$ is connected. Their interior is the first and third quadrants.

Proof by visualization.

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