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Take

$$f(x) = \begin{cases} 1,\quad x∈ℚ \\ 0,\quad x∉ℚ \end{cases}$$

and

$$g(x) = \begin{cases} x,\quad x∈ℚ \\ 0,\quad x∉ℚ \end{cases}$$

where $f(x)$ is not continuous at $a=0$, in fact it's not continuous anywhere, and $g(x)$ is continuous at $a=0$ but not elsewhere.

Now I understand the proof behind $f(x)$ not being continuous anywhere but I don't understand why $g(x)$ is continuous at $a=0$.

I used the logic that $ℚ$ is dense to prove $f(x)$ is not continuous anywhere but given $g(x)$ looks similar, I don't understand how it can be different. I'd appreciate any pointers or tips.

Bill Cogn
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    The continuity of $g$ in $0$ can be directly checked with the usual $\epsilon$-$\delta$-definition. And I'm pretty convinced this question is a duplicate ... –  Jan 07 '24 at 19:39
  • @localshop Actually yes I understand now, looking at the graph $f(x)$ ping pongs between 0 to 1 at $a=0$ and since the rationals and irrationals are dense, it doesn't settle down to 1 value. $g(x)$ does – Bill Cogn Jan 07 '24 at 19:42
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    there is a jump discontinuity at each rational, and the size of the jump is of size of the rational. So the jump at the rational $x=0$ is a jump of size zero, which is no jump at all. – ziggurism Jan 07 '24 at 19:42

1 Answers1

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They are not that similar. In fact, we have $|g(x)|\leq |x|$ for each $x$, so when $|x|$ is small, $g(x)$ is very close to $g(0)=0$. Using this inequality, continuity at $0$ is easy to prove with the $\epsilon-\delta$ definition.

Mark
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