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Let $$ f(x) = \begin{cases} \sin(x) & \text{if $x\in\mathbb{Q}$, and} \\ 0 & \text{otherwise.} \end{cases} $$

I need to study the continuity of the function described, but I don't remember how proceed to solve this.

Xander Henderson
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Mathecm
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    Can you imagine the graph of that function? At what values $x$ would the values $f(t)$ be near $f(x)$ when $t$ is near $x$? – Ethan Bolker Apr 04 '18 at 19:32
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    Related Question: What is the probability that your question will be down-voted? – jim Apr 04 '18 at 19:33
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    I believe this will be continuous at $x$ if and only if $\sin(x) = 0$ – Joe Apr 04 '18 at 19:45
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    I've edited your question so that the question itself is included in the body, rather than just the title (my understanding is that the mobile version of this site makes it difficult to see the title and question body at the same time. That said, I have voted to close this question for "lacking context." What have you tried? What is your definition of continuity? Have you tried to graph it? Have you tried to determine if it is continuous at any particular points (say $x=1$? $x=0$? $x=\pi$? $x=\mathrm{e}$?)? – Xander Henderson Apr 05 '18 at 03:19
  • @XanderHenderson: hopefully your comment will remind Eliakim Machado "how to proceed", suggesting as it does a number of approaches. I believe his context is that he can't remember; having said that, it will be a useful mnemonic to him to read your suggestions. Cheers! – Robert Lewis Apr 05 '18 at 04:00

1 Answers1

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$f(x) \; \text{is continuous at} \; x \Longleftrightarrow x = n\pi, \; n \in \Bbb Z; \tag 1$

for suppose $x \ne n \pi, n \in \Bbb Z$; then

$\sin x \ne 0, \tag 2$

so if $x \in \Bbb Q \setminus \{n\pi \mid n \in \Bbb Z \} = \Bbb Q \setminus \{ 0 \}$ (since $n\pi \in \Bbb Q$ if and only if $n = 0$), then

$f(x) = \sin x \ne 0, \tag 3$

and since there are irrationals $r$ arbitrarily close to $x$ where $f(r) = 0$, we cannot have

$\displaystyle \lim_{y \to x} f(y) = f(x) = \sin x \ne 0; \tag 4$

more formally, set

$\epsilon = \dfrac{\vert \sin x \vert}{2}; \tag 5$

for any $0 < \delta \in \Bbb R$, no matter how small,

$\exists y \; [[\vert y - x \vert < \delta] \wedge [\vert f(y) - f(x) \vert > \epsilon]], \tag 6$

which can't be true if

$\forall \epsilon \exists \delta [\forall y [[\vert y - x \vert < \delta] \Longrightarrow [\vert f(y) - f(x) \vert < \epsilon]]]; \tag 7$

the reader will recognize (7) as a statement of the continuity of $f(x)$ at $x$ thus we have shown that $f(x)$ cannot be continuous at $x \in \Bbb Q \setminus \{0\}$; likewise, if $x \in (\Bbb R \setminus \Bbb Q) \setminus \{n\pi \mid n \in \Bbb Z \}$, that is, a non-rational real which is not of the form $n \pi, n \in \Bbb Z$, then $\sin x \ne 0$ but $f(x) = 0$; there are thus rationals $y$ arbitrarily close to $x$ where $\vert f(y) \vert = \vert \sin y \vert > \vert \sin x \vert / 2$; since $f(x) = 0$, $f(x)$ is not continuous at such $x$.

So we have seen that $f(x)$ is not continuous at any real not of the form $n \pi$, which by contraposition is equivalent to the statement that

$f(x) \; \text{continuous at} \; x \Longrightarrow x = n \pi; \tag 8$

now if $x = n \pi$, then $\sin x = 0$ whether or not $n \pi$ is rational; thus as $y \to n \pi$ through rational values $\sin y \to \sin n \pi$; for irrational $y$ close to $x$, $\sin y = 0 = \sin x$ and in either case $f(y)$ is arbitrarily close to $f(x)$ for $y$ sufficiently close to $x$; thus $f(x)$ is continuous at $n\pi$; this completes the proof of (1).

Robert Lewis
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