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Without using any notion of connected sets how can we show that $\mathbb R$ is the only non-empty clopen set in $\mathbb R$ ? ( actually I am proving that if $f: \mathbb R \to \mathbb R$ is a continuous function satisfying $|x-y| \le M |f(x)-f(y)| , \forall x,y \in \mathbb R $ , for some $M\in \mathbb R$ then $f$ is surjective by showing $f(\mathbb R)$ is clopen )

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marked as duplicate by BCLC, Michael Greinecker Sep 15 '18 at 8:31

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    $\begingroup$ Let $\varnothing \neq U \neq \mathbb{R}$ be open. Show that $U$ is not closed. Consider $x\in U,\, y \in \mathbb{R}\setminus U$ and either $\sup \{ u\in U : u < y\}$ or $\inf \{ u\in U : u > y\}$. $\endgroup$ – Daniel Fischer Jun 25 '14 at 14:19
  • $\begingroup$ Why avoid connectedness? You're basically going to walk through a proof of connectedness without using the word "connected." $\endgroup$ – Neal Jun 25 '14 at 14:22
  • $\begingroup$ @Neal: It's perfectly ok if I have to walk through a proof of connectedness without using the word "connected". Actually I want a basic proof from definitions $\endgroup$ – Souvik Dey Jun 25 '14 at 14:24
  • $\begingroup$ @DanielFischer: Could you please elaborate ? $\endgroup$ – Souvik Dey Jun 25 '14 at 14:37
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Let $\varnothing \neq U \subsetneqq \mathbb{R}$ open. Pick $x\in U$ and $y\in \mathbb{R}\setminus U$. Suppose first that $x < y$. Let

$$z = \sup \{ u \in U : u < y\}.$$

Show that $z \in \overline{U}\setminus U$. Treat the case $y < x$ symmetrically.

Thus you have shown that any nonempty open set other than $\mathbb{R}$ is not closed.

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