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While reading this answer, it puzzles me how can it be that one such a usual function is continuous at 0 and discontinuous in any open interval around zero. Is that correct?

If we take the definition of continuity from the Tom Apostol's Calculus (section 3.3), where a function $f(x)$ is continuous at $p$ if the following implication holds:

$$ \forall \epsilon>0.\exists \delta>0. x\in (p-\delta, p+\delta)\implies \lvert f(x) - f(p) \lvert < \epsilon $$

To me it looks that the following follows:

$$ \forall \epsilon'>0.\exists \delta'>0. \forall x_1, x_2\in (p-\frac{\delta'}{2}, p+\frac{\delta'}{2})\implies \lvert f(x_1) - f(p) \lvert < \epsilon' \land \lvert f(p) - f(x_2) \lvert < \epsilon' \\ \implies -\epsilon' < f(x_1) - f(p) < \epsilon' \land -\epsilon' < f(p) - f(x_2) < \epsilon' $$

when we sum the two inequalities, we get

$$ -2\epsilon' < f(x_1) - f(x_2) < 2\epsilon' \implies \lvert f(x_1) -f(x_2) \lvert<2\epsilon' $$

If we say that $$x_1 = p - \frac{\delta'}{8} \land \epsilon' = \frac{\epsilon}{2} \land \delta=\frac{\delta'}{8} $$

then

$$ \forall \epsilon>0. x\in (x_1-\delta, x_1+\delta)\implies \lvert f(x) - f(x_1) \lvert < \epsilon $$

In other words, $f$ is continuous at $x_1$.

The steps above are not very rigorous, so I am quite sure I did something wrong, I just don't know what. Can someone please point out my mistake?

Thanks!

S11n
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1 Answers1

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To review your proof, given $\epsilon' > 0$ you applied continuity of $f$ at $p$ to first find $\delta' > 0$, and then you used $\delta'$ to find $x_1 = p - \frac{\delta'}{8}$. Fine so far.

Notice: This value of $x_1$ depends on the originally chosen value of $\epsilon'$.

Notice also: You have not yet proved continuity of $f$ at $x_1$.

To prove continuity of the function $f$ at $x_1$, you must restart the whole continuity proof scheme. You cannot just "let" $\epsilon$ be equal to that stale old value of $\epsilon'$ which is sitting around from earlier in the proof. You must start with a brand, spanking new, independently chosen $\epsilon > 0$.

For every value of $\epsilon > 0$, you must find a value of $\delta > 0$ (which will depend on both $x_1$ and $\epsilon$) such that for all $x$ in the domain, $|x - x_1| < \delta \implies |f(x)-f(x_1)| < \epsilon$.

And your proof does not do that.

Lee Mosher
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  • Thanks for the comment and help, professor Mosher! Unfortunately, I still don't understand what exactly I did wrong. I disagree that δ has to be brand new for each ϵ, as long as the implication holds. For example, if it holds for a positive value of epsilon, it will hold for any other larger value of epsilon, for the same delta.
    The first part of my question was focused only on defining the $x_1$ (a value different than $p$), and finding that for all $x_2$ in the neighborhood of $p$, it holds that $ \lvert f(x_1)−f(x_2) \lvert $ can be arbitrarily small. Do we agree about that part?     – S11n Jun 29 '21 at 08:50
  • It is true that you may repeat values of $\delta$, but you should keep in mind that in the end $\delta$ will depend on two quantities, and so it should be a function of those parameter. I rewrote that part of the answer to emphasize those two quantities. In general, like any function with a brand new input value, you should expect the value of $\delta$ to be brand new as, particularly as $\epsilon$ is chosen closer and closer to zero. – Lee Mosher Jun 29 '21 at 14:02
  • Anyway, the exact thing that you did wrong is what I wrote: you cannot just "let" $\epsilon$ be the same as $\epsilon'$. – Lee Mosher Jun 29 '21 at 14:05
  • Here's a good critical tool you can use. Look up, in Apostol and elsewhere, and actual honest to goodness function $f : \mathbb R \to \mathbb R$ which is continuous at $0$ but not at any other number (for example, $f(x)=x^2$ if $x$ is rational, $f(x)=0$ if $x$ is irrational). Then work through your proof, line by line, but using this actual function $f$, and look for where your proof goes wrong. – Lee Mosher Jun 29 '21 at 14:07
  • Thank you for the clarification! In other words, I do not know if for the $x_1$ I defined above I can even find a $\delta$ for each $\epsilon$ smaller than $\epsilon'$. That makes sense. – S11n Jun 30 '21 at 07:19