Surjective function from open interval to closed interval

Question

I am a high school student self-studying analysis, and I came across this question "Does a continuous and surjectve function from $(0,1)$ to $[0, 1]$ exist?"

Of course, because it is hard to intuit uncountability, my intuition suggests that one does not, because $(0, 1)$ is a proper subset of $[0, 1]$. However, thinking about something like $f: (0, 1) \rightarrow \mathbb{R}$ where $f(x) = \text{cot}(\pi x)$ we see that this function is onto the reals. This makes it more intuitive how a function can be onto a codomain that is a superset of its domain.

I understand now that whether a function is injective and surjective can be limited by the cardinalities of the domain and codomain, and whether or not one is a subset of the other is irrelevant.

I want to make sure I am thinking about this correctly. It still intuitively feels wrong to me that this should be possible, and I am wondering if anyone has advice on how to better think about questions of this kind.

Does this answer your question? Continuous bijection from $(0,1)$ to $[0,1]$ — bb_823, Jul 31 '23 at 14:29
Can you find a continuous surjection from $[1/4,3/4]$ to $[0,1]$? If yes, what could you do with the remaining points that keeps the function continuous? — Michael Greinecker, Jul 31 '23 at 14:48
@bb_823 Note that there is a difference betwenn a bijective and surjective function. — Ingix, Jul 31 '23 at 14:56

score 0 · Answer 1 · answered Jul 31 '23 at 15:01

0

Hint: Think about a function defined on $\mathbb R$ like

$$f(x) = \frac{1+\sin(x)}2$$

It's image is the interval $[0,1]$. Can you modify it such that is becomes surjective on $[0,1]$ when the domain is restricted to $(0,1)$?

answered Jul 31 '23 at 15:01

Ingix

14,494

Surjective function from open interval to closed interval

1 Answers1