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I know that this is true for any $z \in X$, but I am unsure of how to prove it.

I was thinking that the best approach would be to demonstrate that $X\backslash{z}$ is path connected, but I'm unsure how to state this in a way that would be reasonable.

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    $\begingroup$ I don't think this is true. $\mathbb{R} \setminus \{0\}$ is not connected. $\endgroup$
    – user301452
    Jun 10, 2016 at 11:16

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You know wrong. Taking $X=\mathbb R$ and $z=0$ proves the statement wrong.

More generally, if $n>1$, you can take

$$X=\{(t,0,0,0\dots,0)| t\in\mathbb R\}\\ z=(0,0,\dots,0)$$

to prove that the statement is false in all dimensions.

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    $\begingroup$ This probably explains why I was finding it difficult to prove, thanks $\endgroup$
    – Andy
    Jun 10, 2016 at 11:28

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