# Proof if $f$ continuous in $x_0$ then there is a neighbourhood of $x_0$ so f bounded.

I have this question :

Proof if $f$ continuous in $x_0$ then there is a neighbourhood of $x_0$ so $f$ bounded.

I want to know if my proof is valid :

If continuous in $x_0$ then :

$$\lim_{x \to x_0} f(x)= f(x_0)$$

Therefore :

All $\epsilon>0$ exists $\delta>0$ so any $x$ that implies $|x-x_0|<\delta$ implies $|f(x)-f(x_0)|<\epsilon$.

Therefore for $\epsilon=1$ there is $\delta>0$ so any $x$ that implies $|x-x_0|<\delta$ implies $|f(x)-f(x_0)|<1$.

$|x-x_0|<\delta \rightarrow x_0-\delta<x<x_0+\delta$.

Lets choose $x_1,x_2$ that implies $x_0-\delta<x_1<x_2<x_0+\delta$

Therefore in $[x_1,x_2]$ from Weierstrass from theorem bounded there.

• fyi, the word you want is "implies" not "appiles". – Simon S Nov 24 '14 at 19:59
• @SimonS Edited. – JaVaPG Nov 24 '14 at 20:01

$$|x-x_0|<\delta \implies |f(x)-f(x_0)|<\epsilon \implies -\epsilon<f(x)-f(x_0)<\epsilon \implies f(x_0)-\epsilon < f(x)<f(x_0)+\epsilon.$$
• You know that $f$ is continuous at $x_0$ but you can't say that it is continuous in $[x_1,x_2].$ Thus, you can't apply Weiertrass. – mfl Nov 24 '14 at 20:52