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Suppose that $f: \mathbb{R}^n \to \mathbb{R}$ is continuous at $P$. Can anyone help me prove that there is an open ball $B$ in $\mathbb{R}^n$ with center $P$ such that $f$ is bounded on $B$.

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What's the definition of continuity you are using? –  Zev Chonoles Apr 28 '12 at 0:17
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Isn't that more-or-less exactly the definition of continuous? Let $\epsilon$ be whatever you like, find $\delta$ and let $B$ have radius $\delta$. Or have I missed something subtle? –  user22805 Apr 28 '12 at 0:34
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$f$ is actually bounded on every open ball (because it is bounded on every closed ball). –  Michael Greinecker Apr 28 '12 at 0:38
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He only stated continuity at a single point, so it is not necessarily bounded on any open ball. @MichaelGreinecker –  Thomas Andrews Apr 28 '12 at 0:45
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What have you tried? If this is homework, tag it as such. –  lhf Apr 28 '12 at 1:12

2 Answers 2

Proof:

Suppose that $f: \mathbb{R}^n \to \mathbb{R}$ is continuous at $P$. Thus, $ \lim_{x\to P}f(x) = f(P)$.

Let $\epsilon = 1 > 0$. Then, there is $\delta > 0$ such that when $x \in \mathbb{R}^n$ and $0 < \| x-P \| < \delta$ then, $|f(x) - f(P)| < 1$.

Then, $B(P, \delta) = $ {$x \in \mathbb{R}^n : \|x - P\| < \delta$ }.

So, if $x \in B(P, \delta)$, then $|f(x) - f(P)| < 1$.

Notice, $|f(x)| - |f(P)| \leq |f(x) - f(P)| < 1$. Hence, $|f(x)| < 1 + |f(P)|$.

Since, $f(P)$ is a constant in $\mathbb{R}$. Let $C = 1 + |f(P)|$. Then, $C \in \mathbb{R}$.

Therefore, we see for all $x \in B(P, \delta)$, we have $|f(x)| < C$.

Thus, there is an open ball $B$ with center $P$ such that $f$ is bounded on $B$.

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That was nicely done. –  ThisIsNotAnId Apr 28 '12 at 15:54

Let's suppose that $f(p)=0$ If it's not, you just take the function $g(x)=f(x)-f(p)$. If f is continuous in $0$, then for every $\epsilon >0$ there is a $\delta >0$ such that, if $|x-0|<\delta$ then $|f(x)-f(0)|<\epsilon$.

That is, $|x|<\delta\rightarrow |f(x)|<\epsilon$

Consider the ball $B$ of center $0$ and the previously obtained $\delta$ radius. Therefore, you have an open ball with center $P=0$ such that $f(x)$ is limited (by $\epsilon$) for every $x$ in it.

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You need to fix a value of $\epsilon$, here. Also, $f$ is continuous at $p$, not at $0$. –  David Mitra Apr 28 '12 at 1:51
    
Instead of $g(x) = f(x) - f(p)$, you should choose $g(x) = f(x-p)$. Then $g$ is continuous at $0$. You may then assume $g(0) = 0$, and then fix $\epsilon > 0$. There exists $\delta > 0$ so that $|x| < \delta$ $\Rightarrow$ $|g(x)| < \epsilon$. So there is an open ball centered at $0$ such that $|g(x)|< \epsilon$ (i.e. $g$ is bounded) on that ball. Unravel what we did, and you get continuity of $f$ at $p$ implies $f$ must be bounded on an open ball about $p$. –  Nicholas Stull Apr 28 '12 at 4:42

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