2
$\begingroup$

This question already has an answer here:

Does there exist a continuous bijection from $\mathbb R$ onto $S^1$ ?

I know that there isn't any continuous bijection from $S^1$ onto $\mathbb R$ because such a continuous bijection would be a homeomorphism , but $S^1$ is compact whereas $\mathbb R$ is not, so impossible. But I'm not sure about this other direction.

Please help.

$\endgroup$

marked as duplicate by Jendrik Stelzner, ℋolo, Noah Schweber, Jyrki Lahtonen, Mark McClure Sep 11 '18 at 14:34

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

0
$\begingroup$

Hint:

Recall the definition of a closed arc segment of a circle $C$. Let

$\quad A_1 \subset A_2 \subset \dots A_k \subset \dots $

be a chain of strictly increasing (expanding at each endpoint) of closed arcs. Then the union of these closed arcs is an open arc of $C$.

$\endgroup$
0
$\begingroup$

There is no such map. It suffices to show that there is no bijective map $(0,1) \to S^1$.

If there were, then it would restrict to a bijection $f:(0,1) \setminus \{pt\} \to S^1 \setminus \{f(pt)\}$.

In other words there would be a continuous bijection $f:I_1 \coprod I_2 \to I$, where $I_k$ are intervals, and $I$ is also an interval.

I claim that any such map would be an open map and hence a homeomorphism. To see this, we need only show that $f(I_k)$ is an open interval. Note that $f(I_k)$ is a connected component of $\mathbb R$, so of the form $[a,b),[a,b],(a,b],(a,b)$.

Suppose that it is closed on the left. Take a preimage $f^{-1}(a)$ and note that since $I_k$ is open there exists an open ball containing the preimage, say $(x,y)$. Then the argument is basically examine $f(x)$ and $f(y)$ and apply the intermediate value theorem since by hypothesis these are different points.

Clearly, $f$ cannot be a homeomorphism since $I$ is connected.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.