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This is my proof

Rewriting the proof (it's not very good)

Proof

Since $f$ is continuous, we have $\forall \epsilon > 0$, $\exists r > 0$ s.t.

$|f(x) - f(y)| < \epsilon$ whenever $\Vert x - y\Vert < r$

We are also told that $f(x) < C$ for some $C \in \mathbb{R}$

So if I go with the suggest that to choose $\epsilon = C - f(x)$, then I get

$|f(x) - f(y)| < C - f(x) \implies f(x) - C < f(x) - f(y) < C - f(x) \implies -C < -f(y) < C - 2f(x) \implies C - 2f(x) < f(y) < C$

Problem I have here is that I didn't use $r$ at all. I know I am doing something wrong, could someone point it out to me?

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    How exactly does one know to pick $\epsilon = C - f(x)$ here?2012-11-11

2 Answers 2

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So,

  1. $C$ can also be negative, it is not said anywhere and isn't needed to be restricted.
  2. I would denote things clearer, for example, it's ok to denote the Euclidean norm in $\Bbb R^n$ by $||.||$, but for $\Bbb R$, I would rather use $|.|$.
  3. As Stefan commented, this $f(x) doesn't always imply $|f(x)-f(y)|<|C-f(y)|$. And how did you get the continuation: $|C-f(y)|<||x-y||$?

I suggest you to start it over, use the very definition of continuity, and use the positive number $\varepsilon:=C-f(x)$.

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    Please see edit2012-11-10
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I think your first step ($\forall C>0$) is not correct. Since you write in your question "topology" I assume that you know some basic topological facts. Since $f$ is continuous, therefore $f^{-1}$ is an open map. Now the set $(-\infty,C)$ is open so $f^{-1}((-\infty,C))$ is also open. It implies that there exists $r>0$ such that $f(B_r(\textbf{x}))\subset(-\infty,C)$.

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    No I should've deleted that. I was supposed to have written "continuity".2012-11-10