Is the interior of a connected set in $\mathbb R^k$ connected?
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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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Nope... Pick two tangent balls... |
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No. Let $A_1=\{(x,y)|x\leq0,y\leq0\}$ (The fourth 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 fourth quadrants. Proof by visualization. |
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