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The standard proofs can be found here: Every function $f: \mathbb{N} \to \mathbb{R}$ is continuous?

But I want to see how this could be proved using directly the definition of continuity of real functions:

$ \forall x,y \;\forall \epsilon > 0 \; \exists \delta > 0 $ such that $ |x-y|<\delta \implies |f(x)-f(y)|<\epsilon$

I tried $\delta$ = 1 and $\delta=\frac{\epsilon}{2}$ but didn't work for me. Thanks.

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Actually, $\delta=1$ works. For natural numbers, $|x-y|<1$ means that $x=y$ and $f(x) = f(y)$, so $|f(x)-f(y)|=0<\epsilon$.

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  • $\begingroup$ Oh!! So obvious and perfect. Thank you very much. $\endgroup$ – user286485 Nov 10 '15 at 21:22
  • $\begingroup$ What if $f(x) = \frac{1}{x-y}$, for some $y \in \mathbb{N}$? $\endgroup$ – the_candyman Nov 10 '15 at 21:22
  • $\begingroup$ @the_candyman it is a function $\mathbb{N}\setminus\{y\}\rightarrow\mathbb{R}$, not $\mathbb{N}\rightarrow\mathbb{R}$, since it is not defined in $y$. $\endgroup$ – lisyarus Nov 10 '15 at 21:24
  • $\begingroup$ @lisyarus ok, got it, thanks! $\endgroup$ – the_candyman Nov 10 '15 at 21:25

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