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>> No.11433578 [View]
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11433578

Need a bit of help with a topology task:
Prove that the closure [math]\bar{A}[/math] of any connected set [math]A[/math] is connected.

I found an easy prove that for any [math]x \in \bar{A} \setminus A[/math], [math] A \cup \{x\} [/math] is also connected.
My question is if that's enough, or if that only proves you could add a finite number of points to A and have it still be connected.
It feels like that, because you're only adding a point at a time, but then again you could say do this for all [math]x \in \bar{A} \setminus A[/math].

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