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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.)2012-05-05

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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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    @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.2013-07-03
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Nope... Pick two tangent balls...

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    BTW: probably a better example is two non-touching disks connected by a segment o$f$ line.2012-05-06
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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.