Let the set $S^1:=\{x\in\Bbb R^2:\|x\|_2=1\}$ but with the topology not induced from the euclidean norm (the euclidean norm is just used here to define the elements of the set), instead the topology is induced from the discrete metric defined as

$$d(x,y)=\begin{cases}0,&\text{ if }x=y\\1,&\text{ if }x\neq y\end{cases}$$

Now the set $[0,1]$ have the same topology induced from the discrete metric. We can say in this case that $S^1$ and $[0,1]$ are homeomorphic?

I think that the answer is yes. Observe that any bijective function between the sets is continuous because every subset of any of these spaces is open and closed.

Because the sets $S^1$ and $[0,1]$ have the same cardinality then a bijection exists, then we conclude that they are homeomorphic. It is this reasoning correct?

  • $\begingroup$ Yes. My only objection is to the phrase "bijection between the sets". Strictly speaking, a function is always from one set to another. $\endgroup$ – DanielWainfleet Nov 23 '16 at 9:05
  • $\begingroup$ @user254665 well, but In my experience a function can be extended, restricted or induced in some other sets, so I think a bit of "redundancy" is not too bad. $\endgroup$ – Masacroso Nov 23 '16 at 9:50

Your argument is correct.

More precisely, if $f: S^1 \to [0, 1]$ is a bijection, then for every (open) set $M \subset [0, 1]$ the preimage $f^{-1}(M)$ is open (since every subset of $S^1$ is open in the discrete topology). Conversely, for every (open) set $N \subset S^1$ the preimage $(f^{-1})^{-1}(N)$ is open in $[0, 1]$.

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  • $\begingroup$ Ah, thank you, I needed the confirmation to be totally sure. $\endgroup$ – Masacroso Nov 22 '16 at 11:45

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