0

I am trying to prove that there is no continuous bijective function from $ \{ 0,1\} ^\mathbb{N}$ to $[0,1]$ where $\{0,1\}$ has the discrete topology, $\{ 0,1 \}^\mathbb{N}$ has the product topology and $[0,1]$ has the usual topology (order topology).

I don't know how to start, I tried by contradiction, but I don't really see the contradiction. Could someone give me a hint? I feel like it's something simple but I really don't see it.

Alejandro Peralta
  • 71
  • 1
    Hint: prove first that these spaces are not homeomorphic. – Moishe Kohan Sep 13 '23 at 03:44
  • 1
    Hint: If you have a continuous bijection from a compact space to a Hausdorff space, what can you say about its inverse? (recall closed subsets of compact spaces are compact, and compact subsets of hausdorff spaces are closed) – M W Sep 13 '23 at 03:45
  • Hmmm, maybe the combination of these two hints sort of gave the game away:) – M W Sep 13 '23 at 03:50
  • @MW Thank you, that helps a lot. I think I can use connectedness to show that they are not homeomorphic. Is ${ 0,1 }^\mathbb{N}\setminus {a}$ connected? – Alejandro Peralta Sep 13 '23 at 04:06
  • 1
    @AlejandroPeralta As User's answer explains, its totally disconnected, even without removing a point. Also another (more or less equivalent) route would be to say $f^{-1}$ can't be continuous since then composing with projection onto first coordinate would give a continuous surjection from $[0,1]$ to ${0,1}$. Then you don't really think about the product topology as much, other than the fact projections onto coordinates are continuous. – M W Sep 13 '23 at 04:14

1 Answers1

2

MW gives in their comment half of the solution: if you have a continuous bijection $f$ from a compact space to a Hausdorff space, then $f$ is a homeomorphism. We have that $\{0,1\}^{\Bbb N}$ is compact by Tychonoff's theorem.

If there's a continuous surjection $f: [0,1] \to \{0,1\}^{\Bbb N}$, then $\{0,1\}^{\Bbb N} = f([0,1])$ is connected. However, since $\{0,1\}$ is totally disconnected (discrete topology), then so is $\{0,1\}^{\mathbb N}$. This is a contradiction.

User
  • 822
  • 2
  • 10