Is there a continuous function from $[0, 1]$ onto $(0, 1)$?

Question

If there is none, why?

And for the other side, what about open set $(0, 1)$ to closed set $[0, 1]$ with a continuous function?

Thanks

@learner: I disagree with this edit (and would have rejected it myself), the readability of the post did not improve, and all the effect was bumping up a question from two and a half years ago, instead of letting the newer questions (and lord knows, we don't have a shortage of those) to be seen on the front page. Every edit bumps the question, so bump with care! — Asaf Karagila, Apr 08 '13 at 12:28
@AsafKaragila: The system already discriminates so strongly in favour of the youngest among questions, which aren't necessarily the most interesting ones, that I don't mind if an older question returns to the front page now and then. Let's say, if I find an annoying typo in a question or answer, even if it is one of my own, I won't hesitate to edit it away. I have no particular opinion about the current question though. But replacing non-math by a math formula is almost always an improvement. — Marc van Leeuwen, Apr 08 '13 at 13:05
@Marc: Typos are one thing; adding $ symbols to math which wasn't any less readable than it is now (say, compared to things like cos x=sin x) is really just pigging out on the main page resources. — Asaf Karagila, Apr 08 '13 at 13:09

score 12 · Accepted Answer · answered Nov 09 '10 at 16:44

12

HINT: For the first one use the fact that, Continuous image of a compact set is compact.

answered Nov 09 '10 at 16:44

Thanks. But for the other side, can a non-compact set be mapped to compact? Actually I know a little about topology, but I cannot find info about mapping open set to closed. – haohaolee Nov 09 '10 at 16:48
@haohaolee: Thinking! – Nov 09 '10 at 16:48
Yes. I can even figure out some examples, e.g., sinx, but it would be better if I can get the serious proof – haohaolee Nov 09 '10 at 16:54
@haohaolee: Myke did my job! – Nov 09 '10 at 16:57

score 9 · Answer 2 · answered Nov 09 '10 at 16:53

9

For the other side consider $f: (0,1) \to [0,1]$ defined as $f(x)= |\cos(2\pi x)|^{2}$

answered Nov 09 '10 at 16:53

Mykie

3

Perhaps mapping $[1/3,2/3]$ linearly to $[0,1]$ and mapping $(0,1/3)$ to $f(1/3)$ and $(2/3,1)$ to $f(2/3)$ would be a simpler way to see there is an example. – Patrick Da Silva Dec 05 '11 at 07:59
Why did you square $|\cos(2 \pi x)|$? Without the square, isn't the range the same? – SRobertJames Dec 05 '22 at 02:26

2 Answers2