Timeline for A problem in real analysis of a topological nature

7 events

when toggle format	what		by	license	comment
Dec 27, 2018 at 15:49	comment	added	Nate Eldredge		@მამუკაჯიბლაძე: Yes, thank you. Fixed.
Dec 27, 2018 at 15:49	history	edited	Nate Eldredge	CC BY-SA 4.0	closure
Dec 26, 2018 at 17:24	comment	added	Wojowu		You are right, I was a little too quick there. We can either use jump discontinuity as you mention, or we can observe $(q,f(q))$ is an isolated point of the graph.
Dec 26, 2018 at 17:16	comment	added	Nate Eldredge		@Wojowu: It's not quite that simple, I think. For instance, the function $f=1_\mathbb{Q}$ is discontinuous at every rational (and every irrational), but its restriction to $S=\mathbb{Q}$ is uniformly continuous. So the fact that $f$ is discontinuous at the rationals doesn't immediately rule out the possibility for $S$ to contain rationals. We would have to work a little harder. I guess the key is that $f$ has a "jump" discontinuity at every rational.
Dec 26, 2018 at 17:09	vote	accept	James Baxter
Dec 26, 2018 at 17:08	comment	added	Wojowu		Slight simplification of the proof: $S$ cannot contain a single rational number, since $f$ is discontinuous at every rational, hence so is $f\|_S$ for any dense $S$. Thus $S\subseteq\mathbb R\setminus\mathbb Q$, but then we have problems around $0$.
Dec 26, 2018 at 17:01	history	answered	Nate Eldredge	CC BY-SA 4.0