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The title is lacking because the whole question didn't fit: For a topological space $X$ with at least two points $X$ such that there exists a continuous bijection $f: X \rightarrow \mathbb{R}$, prove there exists $x$ s.t. $X \setminus \{x\}$ isn't connected.

We know there are at least two points in $X$, labeled $x_1,x_2$. We can label $f(x_1) = y_1$,$f(x_2) = y_2$, and by continuity, for an open interval $B_r$, $f^{-1}(B_r(y_1), f^{-1}(B_r(y_2) $ are both open. I'm not sure how to continue from here.

I can't even convince myself that the question is indeed true, for example, taking $X = \mathbb{R}, f(x) = x$. I can't convince myself that $\mathbb{R} \setminus \{x_0\}$ isn't connected. Is it really the union of two disjoint closed sets?

Any help would be appreciated.

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    $\begingroup$ To answer the last question; yes. $(-\infty,x_0)$ is open and closed in the subspace topology (which is what you are interested in). Therefore $\Bbb R\setminus\{x_0\}$ is not connected. $\endgroup$ – Clayton Sep 13 '15 at 19:04
  • $\begingroup$ Isn't $x_0$ a limit point? Shouldn't all limit points be part of a closed set? $\endgroup$ – John H. Sep 13 '15 at 19:08
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    $\begingroup$ If you are looking at it as a subset of $\Bbb R$, yes. But if we put $Y=(-\infty,x_0)\cup(x_0,\infty)$, then no. $\endgroup$ – Clayton Sep 13 '15 at 19:09
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$\Bbb R\setminus\{0\}=(-\infty,0)\cup(0,\infty)$. The point here is that $(-\infty,0)$ and $(0,\infty)$ in the subspace topology of $\Bbb R\setminus\{0\}$.

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