In which of the following cases is there no continuous function from the set $S$ onto the set $T$?

Question

In which of the following cases is there no continuous function $f$ from the set $S$ onto the set $T$ ?

a. $S =[0,1], T = \mathbb R$

b. $S = (0,1), T = \mathbb R$

c. $S = (0,1), T = (0,1)$

d. $S = \mathbb R, T = (0,1)$

(a) is correct, since the continuous image of a compact set is compact, and $\mathbb R$ is not compact.

(c) is incorrect, since the continuous preimage of an open set is open.

Now in (d) I'm confused. $\mathbb R$ is both open and closed, so if we treat $\mathbb R$ in (d) as an open set, then (d) should be incorrect.

Am I right?

Answer 1

$f(x)=\tan(-\frac {\pi} 2+{\pi} x)$ is a homeomorphism from $(0,1)$ to $\mathbb R$. Does this give you the answer to b) and d)? The correct argument for c) is to say that the identity map is a continuous map. Your argument is not correct.

Answer 2

(a) This is the correct choice. There is no continuous function from [0,1] onto $\mathbb R$, because [0,1] is compact and $\mathbb R$ is not, as you said.

(b) Incorrect that there is no continuous function. $f(x)=$ ${{2x-1}}\over{x(x-1)}$ is a continuous function from (0,1) onto $\mathbb R$.

(c) Incorrect that there is no continuous function. $f(x)=x$ is a continuous function from (0,1) onto itself.

(d) Incorrect that there is no continuous function. Inverse of (b).