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Aside from rigor, is this proof correct?

Claim. Let $f$ be a function defined on $[0, 1]$ such that $\lim\limits_{y\to a} f(y)$ exists for all $a \in [0, 1]$. Then for any $\epsilon > 0$ there are only finitely many points $a \in [0, 1]$ with $$|\lim\limits_{y\to a} f(y) - f(a)| > \epsilon.$$

Proof. Suppose that there are infinitely many such points $a.$ Then by the Bolzano-Weierstrass Theorem, these points have a limit $x \in [0, 1].$ Let $$L := \lim\limits_{y \to x} f(y) = \lim\limits_{a\to x} f(a).$$

The condition $$|\lim\limits_{y\to a} f(y) - f(a)| > \epsilon$$ means that for $y$ close to $a$, $f(y)$ is far from $f(a).$ Similarly $\lim\limits_{a \to x} f(a) = L$ means that for $a$ close to $x,$ $f(a)$ is close to $L.$ Together this means that for $y$ close to $x$, $f(y)$ is far from $L$, but this contradicts the fact that $L = \lim\limits_{y \to x} f(y),$ i.e. for $y$ close to $x$, $f(y)$ is close to $L.$

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    $\begingroup$ That is a funny first-liner! $\endgroup$
    – Michael
    Jul 7 '16 at 2:53
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    $\begingroup$ Seems like you are on the right track though. Another approach would be to use the "reverse triangle inequality" $|c-d| \geq |c-w| - |d-w|$ via $$ |f(a_n^*)-L| \geq |f(a_n^*)-f(a_n)| - |f(a_n)-L| $$ where $\{a_n\}_{n=1}^{\infty}$ is the infinite sequence you get from Bolzano-Wierstrass (which converges to $x$), and $a_n^*$ is another sequence that you define in a convenient way. $\endgroup$
    – Michael
    Jul 7 '16 at 3:10
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    $\begingroup$ Ah I see. So for each $a_n$ we choose an $a^*_n$ close to it so that $|f(a_n^*)-f(a_n)|$ is big. Since $|f(a_n)-L|$ is small, the triangle inequality says $|f(a_n^*)-L|$ is big. Nice. $\endgroup$
    – trong
    Jul 7 '16 at 4:03
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    $\begingroup$ Yes, just choose each $a_n^*$ to be "sufficiently close" to $a_n$ so that $|f(a_n^*)-f(a_n)|\geq \epsilon$. The only other thing is that you also need the distances between $a_n^*$ and $a_n$ to be converging to 0 so that $a_n^*$ also approaches $x$. $\endgroup$
    – Michael
    Jul 7 '16 at 14:30
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This proof is missing more than just "rigor"--there is some real subtlety in making the estimates you make at the end precise. In particular, it is not necessarily true that $f(y)$ is far from $L$ for all values of $y$ sufficiently close to $x$. Fortunately, in order to contradict the fact that $L = \lim\limits_{y \to x} f(y)$, you only have to show that $f(y)$ is far from $L$ for some values of $y$ arbitrarily close to $x$.

Let's take a closer look at your claims. For any fixed $a$, $|\lim\limits_{y\to a} f(y) - f(a)| > \epsilon$ implies that there exists $\delta_a>0$ (possibly depending on $a$) such that $|f(y)-f(a)|>2\epsilon/3$ when $0<|y-a|<\delta_a$. Also, $\lim\limits_{a \to x} f(a) = L$ implies that there exists a $\delta>0$ such that $|f(a)-L|<\epsilon/3$ whenever $0<|a-x|<\delta$. Combining these two facts, you can conclude that whenever $0<|a-x|<\delta$ (and $|\lim\limits_{y\to a} f(y) - f(a)| > \epsilon$) and $0<|y-a|<\delta_a$, $|f(y)-L|>\epsilon/3$.

However, it is not necessarily true that every $y$ which is sufficiently close to $x$ satisfies $0<|y-a|<\delta_a$ for some $a$ such that $|\lim\limits_{y\to a} f(y) - f(a)| > \epsilon$ and $0<|a-x|<\delta$. So you can't claim that $f(y)$ is far from $L$ for all $y$ sufficiently close to $x$. What you can say is that for any $\delta'>0$, there exists $y$ such that $0<|y-x|<\delta'$ and $|f(y)-L|>\epsilon/3$: just choose $a$ such that $0<|a-x|<\min(\delta'/2,\delta)$ and $y$ such that $0<|y-a|<\min(\delta'/2,\delta_a)$. This is then enough to reach a contradiction.

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  • $\begingroup$ Ah, you're right. It's not true for every $y$, e.g. one of the other $a$'s would have $f(a)$ close to $L$. In the same line I think you meant $|f(y) - f(a)| > \epsilon$ instead of $|\lim\limits_{y\to a} f(y) - f(a)| > \epsilon$? Thanks for the detailed answer! $\endgroup$
    – trong
    Jul 7 '16 at 3:43

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