# Topological definition of continuity on a real valued function with removable discontinuity

Consider some constant function $f(x)=c$, $f(x_0)=0$, for $c\neq 0$. This function is obviously discontinuous as $x_0$, so according to the topological definition of continuity, there must exist an open set $U\subset \mathbb{R}$ such that $f^{-1}(U)$ is not open. I'm having trouble finding such a set.

Naturally, I would want to consider the interval $(c-\epsilon, c+\epsilon)$ for some $\epsilon<c$, but the preimage of this is $(-\infty, x_0)\cup (x_0, \infty)$, which is open.

Thanks so much!

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You mean "there must exist an open set $U\subseteq\mathbf{R}$ such that $f^{-1}(U)$ is not open." –  Keenan Kidwell Nov 15 '12 at 18:24
Quite right! Thank you for that. –  William Stagner Nov 15 '12 at 18:55
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## 1 Answer

Try looking at a small $\epsilon$-neighborhood around $0$ - its preimage will only contain the singleton $\{x_0\}$.

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Ahh, that makes sense! What if the function is simply not defined at $x_0$? –  William Stagner Nov 15 '12 at 17:23
Oh, then it would be continuous on it's domain. –  William Stagner Nov 15 '12 at 17:26
That's correct. The definition of continuity requires the function to be defined. –  icurays1 Nov 15 '12 at 17:28
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