Currently I am trying to prove that 1/3-Cantor set is homeomorphic to $\{0,1\}^\mathbb{N}$ I tried personal approach but after a while, when I got stuck I went with research and found this and especially this. So in the second post everything seems clear to me, to the point where author intersects C-set with interval $[h^*,h^*+(2.3)^{-j})$. I see from where it came, but I am not quite sure why it is 2.3. I mean, I see that we are looking at bare construction of Cantor set and we start "building" its parts starting from point 0, but I don't see why it has to be by 2,3. Or maybe it's some mistake in notation... idk.

Still, despite it, I also tried to check his case for a=1, which is meant to be similar. And I would be grateful if someone can check it and say if whether it's wrong or not.


For $ a=1, j \in N, h \in H(j,1)$ we have:

$ \{x\in C:\forall i\leq j (x_i=h(i)\}= C\cap (h^*-2.3^{-j},h^*] $ as we go from point 1.

Now h* is either 1 or lower endpoint of $(h^*,h^*+3^{-j+1})$, which were deleted due to construction of C. Moreover because a=1 we have

$ \{x\in C:\forall i\leq j (x_i=h(i)\}= C\cap (h^*-2.3^{-j},h^*+3^{-j+1}) $ which is open in C.

It may be trivial, but still, I will be grateful if someone can look at it and point out my mistakes. Thanks for help!

  • 1
    $\begingroup$ Looks OK to me. I think an easier way to see that $X=\{0,1\}^N$ and $C $are homeomorphic is to show that $S\subset X$ is closed in $X$ iff $\{t^*:t\in S\}$ is closed in $C$ : We have $s\in \bar S$ iff there is a sequence $(s_j)_{j\in N}$ in $S$ that converges pointwise to $s.$...( That is $\lim_{j\to \infty}s_j(n)=s(n)$ for each $n\in N.$)... And $(s_j)_{j\in N}$ converges pointwise to $s$ iff $(s_j^*)$ converges to $s^*.$ (Notation as in my linked answer). (Note: $(s_j(n))_{j\in N}$ is a binary sequence so it converges iff it is eventually constant.) $\endgroup$ – DanielWainfleet Aug 23 '16 at 23:32
  • $\begingroup$ @user254665 Great thanks! And I didn't expected reply from author himself :D Also I'll try approach you sugested in comment, as exercise. Anyway... thanks! $\endgroup$ – Kiwi Aug 24 '16 at 7:53

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