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What's wrong with my following proof?

Suppose $f:\Bbb R\to \Bbb R$ is continuous. Take $x,y\in \Bbb R$.

Given $\epsilon >0$ there exists $\delta_1,\delta_2$ such that $$ 0<|x-x_0|<\delta_1 \rightarrow |f(x)-f(x_0)|<\epsilon\\ 0<|y-x_0|<\delta_2\rightarrow |f(y)-f(x_0)|<\epsilon $$ For some fixed $x_0$.

Then $|f(x)-f(y)|\leq |f(x)-f(x_0)|+|f(y)-f(x_0)|<2\epsilon$ provided $|x-y|<\min\{\delta_1,\delta_2\}$.

So we've found a $\delta$ which works for all $x,y$. This is obviously wrong somewhere, but where?

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    $\begingroup$ The assumption $|x-y|<\min\{\delta_1, \delta_2\}$ does not imply that $|x-x_0|<\delta_1$ and/or $|y-x_0|<\delta_2$. $\endgroup$ Commented Mar 1, 2016 at 18:44

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The $\delta$ you found only works for $x$ and $y$ that are close to $x_0$, so choosing a different $x_0$ means you might need a different $\delta$.

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It has to work for every $x_0$! Who would be $x,y$ in the definition of uniform continuity?

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  • $\begingroup$ I'm using the $x,y$ as the ones in the part $\forall\epsilon\,\exists \delta \,\forall x\,\forall y (|x-y|<\delta \cdots)$, so it shouldn't matter if I fix some $x_0$ in between. $\endgroup$ Commented Mar 1, 2016 at 18:46
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    $\begingroup$ But your value of $\delta$ depends, potentially, on $x_0$, which it must not do for uniform continuity. Try your approach on $1/x$ over $(0, \infty)$. $\endgroup$
    – Brian Tung
    Commented Mar 1, 2016 at 18:53
  • $\begingroup$ @BrianTung Yes $\delta=\delta(x_0)$, but I claim that this does not matter because we're only picking one $x_0$ (therefore just a single $\delta$), what's wrong with this? $\endgroup$ Commented Mar 1, 2016 at 19:04
  • $\begingroup$ @user1293038: Consider the function I mentioned: $1/x$ over $(0, \infty)$. What value of $x_0$ are you going to choose? Suppose $\varepsilon = 1/10$. (Uniform continuity requires that $\delta$ depend only on $\varepsilon$, not on $x$.) What value of $\delta$ will you produce? Will that $\delta$ work for $x = 1/10$? For $x = 1/100$? For $x = 1/1000$? Etc. $\endgroup$
    – Brian Tung
    Commented Mar 1, 2016 at 19:12
  • $\begingroup$ This comment is not quite clear. Could you clarify? $\endgroup$ Commented Mar 1, 2016 at 19:37
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Uniform continuity requires you to find, for any given $\varepsilon > 0$, a $\delta$ that works for the entire domain of $f(x)$: The $\delta$-neighborhood of any $x$ maps into the $\varepsilon$-neighborhood of $f(x)$.

That is not what you've shown. You've (more or less) shown that for any given $\varepsilon > 0$ and any given $x$, you can find a $\delta$-neighborhood of $x$ such that any two points in that neighborhood map to points that are within $\varepsilon$ of each other. That isn't really any different from bog-standard continuity.

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