Continuity implies boundedness in an interval proof

[tylerc1991]

Homework Statement

If a function $f$ is continuous at a point $x$, then $f$ is bounded on some interval centered at $x$. That is, $\exists M \geq 0$ s.t. $\forall y$, if $|x - y| < \delta$, then $|f(y)| \leq M$

Homework Equations

The Attempt at a Solution

Let $\varepsilon > 0$. Since $f$ is continuous at $x$, $\exists \delta > 0$ s.t. $\forall y$, if $|x - y| < \delta$, then $|f(x) - f(y)| < \varepsilon$. Now,

$|f(x) - f(y)| < \varepsilon \iff$

$- \varepsilon < f(x) - f(y) < \varepsilon \iff$

$f(x) - \varepsilon < f(y) < f(x) + \varepsilon$.

Stated differently,

$f(y) \in (f(x) - \varepsilon, f(x) + \varepsilon)$.

We are trying to find an $M \geq 0$ s.t. $|f(y)| < M$, or $f(y) \in (-M, M)$. This is where I get a little stuck. I realize that we may choose any $\mu > 0$ and say that $M = f(x) + \varepsilon + \mu$. But I run into trouble when doing this. I want something like $M = f(x) \pm (\varepsilon + \mu)$. This would cover the interval that I am trying to get $M$ into, but I don't exactly know how to say it (except by the way just mentioned of course). Is there a way to write this as a SINGLE value (as opposed to the $\pm$ showing up)?

EDIT: What about $M = f(x) + |\varepsilon + \mu|$?

Thank you for your help!

 

[micromass]

Can't you just do

$$M=\max\{f(x)+\varepsilon,-f(x)+\varepsilon\}$$

I see no use for the $\mu$...

 

[Petek]

If you have to prove that a function is continuous, you have to show that certain conditions are satisfied for every value of $\epsilon > 0$. However, if you're given that a function is continuous and you have to prove something about it, you can select a specific, convenient value for $\epsilon > 0$. So let $\epsilon = 1$. Since f is continuous at x, $\exists \delta > 0$ such that if $|x - y| < \delta$, then $|f(x) - f(y)| < 1$. Can you take it from there?

 

[tylerc1991]

Can't you just do
$$M=\max\{f(x)+\varepsilon,-f(x)+\varepsilon\}$$
[/itex]...

Since $M \geq 0$, where is the negative $f(x)$ coming from and how would it fit into choosing an $M$?

... if you're given that a function is continuous and you have to prove something about it, you can select a specific, convenient value for $\epsilon > 0$. So let $\epsilon = 1$. Since f is continuous at x, $\exists \delta > 0$ such that if $|x - y| < \delta$, then $|f(x) - f(y)| < 1$. Can you take it from there?

Right, but I could just plug in $\varepsilon = 1$ into what I had written so far and I would still be in the same boat (unless I am just completely misunderstanding). I am getting stuck on picking an $M$ that will work. I can see how to do it if everything is positive ($M = f(x) + \varepsilon$), but I don't see how to handle it if it starts to dip into the negatives.

 

[Petek]

An additional hint:

$$|f(y)| = |f(y) - f(x) + f(x)| \leq |f(y) - f(x)| + |f(x)|$$

 

[micromass]

Since $M \geq 0$, where is the negative $f(x)$ coming from and how would it fit into choosing an $M$?

The negative f(x) is to be able to deal with negative function values.

If $|f(y)|\leq -f(x)+\varepsilon$, then $f(x)-\varepsilon\leq f(y)$. So the f(y) is bounded below.