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$C$ is cantor set on $[0,1]$.

[Definition: $A_0$ is the interval $[0,1]$. $$A_n=A_{n-1}-\bigcup_{k=0}^{\infty}\Big(\frac{1+3k}{3^n},\frac{2+3k}{3^n}\Big),\space\space n=1,2,\dots$$ The intersection $$C=\bigcap_{n=0}^{\infty} A_n$$ is called Cantor set on $[0,1]$]

(a) Show that $C$ is totally disconnected.

<p>(b) Show that $C$ is compact.</p>

<p>(c) Show that each set $A_n$ is union of finitely many disjoint closed interval of length $1/3^n$, end show that the end points of this interval lie in $C$.</p>

<p>(d) Show that $C$ has no isolated points.</p>

<p>(e) Conclude that $C$ is uncountable.</p>

I have no problem of showing the first four ((a), (b), (c) and (d)) part of the proof. But got stuck at (e). I know this theorem:

Let $X$ be the nonempty compact Hausdorff space. If $X$ has no isolated points, then $X$ is uncountable.

How can I show that $C$ is Hausdorff subspace of $[0,1]$?

topology_001
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  • You should show Cantor set is perfect. Then use this: https://math.stackexchange.com/questions/201922/proof-that-a-perfect-set-is-uncountable – GhD Nov 05 '17 at 12:37
  • Thanksssss a lotttttt for the link @GhD I know the solution now for (e). – topology_001 Nov 05 '17 at 12:40

1 Answers1

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Any subspace of a Hausdorff space is Hausdorff! $[0,1]$ is Hausdorff, so there you go.

M. Van
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  • umm, yeah! that should be the case (I know!)... [while writing this comment, just the solution bumped my head]. Thanks a lot, I know the answer now! – topology_001 Nov 05 '17 at 12:37
  • @M.Van [0,1] is connected,,,,my question is that how [0,1] is hausdorff ? – jasmine Apr 13 '18 at 10:50
  • any metric space is hausdorff, [0,1] is a metric space with the absolute value. (to show a metric space is Hausdorff, consider $x,y$ distinct, and let $U_x=B(x, d(x,y)/2))$, $U_y=B(y,d(x,y)/2))$. Now for a point $z$ that is in both sets we'd have $d(x,y) \leq d(x,z)+d(y,z) < d(x,y)$, contradiction, so these open sers are disjoint) – M. Van Apr 14 '18 at 16:11