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.

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

Rephrasing your argument:

$$|x-x_0|<\delta \implies |f(x)-f(x_0)|<\epsilon \implies -\epsilon<f(x)-f(x_0)<\epsilon \implies f(x_0)-\epsilon < f(x)<f(x_0)+\epsilon.$$

| cite | improve this answer | |
  • $\begingroup$ Oh, I understand your proof is much easier, I wonder if the way I proofed it correct?, because I had an exam today, and that's how I wrote it. $\endgroup$ – JaVaPG Nov 24 '14 at 20:05
  • 1
    $\begingroup$ 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. $\endgroup$ – mfl Nov 24 '14 at 20:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.