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This is my proof

Rewriting the proof (it's not very good)


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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$\lVert f(x) - f(y) \rVert < \lVert C - f(y)\rVert$ doesn't hold, i.e. $f(x) = 1, f(y) = (2)$, then $\lVert f(x)-f(y) \rVert = \lvert 1-2 \rvert = 1$ but $f(x)<2$ and $\lvert 2-2 \rvert = 0$. – Stefan Nov 10 '12 at 20:20
Perhaps I should employ triangle inequality? – Hawk Nov 10 '12 at 20:22
Your proof is now mostly correct, except that you forgot to flip the sign on $C-2f(x)$ in the last step. You don't need to "use" $r$, you should just argue more explicitly that by the continuity condition quoted above, there is an $r$ for which this chain of inequalities holds. ($\lVert x-y\rVert\lt r$ is equivalent to $y\in B_r(x)$.) – joriki Nov 11 '12 at 10:41
How exactly does one know to pick $\epsilon = C - f(x)$ here? – Hawk Nov 11 '12 at 18:12
up vote 3 down vote accepted


  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 $f(x)<C$ 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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But $f(x)$ isn't a number? – Hawk Nov 10 '12 at 21:01
$f(x)$ is a number, as $f: \mathbb R ^n \to \mathbb R$. – Stefan Nov 10 '12 at 21:11
Please see edit – Hawk Nov 10 '12 at 21:47

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". – Hawk Nov 10 '12 at 20:55

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