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prove that the ternary cantor set is compact and a perfect set.

My trial:

I know that I should prove that it is closed and bounded, for proving that it is closed because finite union of closed sets (the cantor set definition is this ) are closed and also using that arbitrary intersection of closed sets are closed.

Now to prove that it is bounded shall I prove that its lenght is zero?

How can I prove that it has no isolated points?

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    $\begingroup$ It's contained in the unit interval. $\endgroup$ Mar 12 '19 at 6:34
  • $\begingroup$ @LordSharktheUnknown thank you .... so I will edit my question to another one $\endgroup$
    – Secretly
    Mar 12 '19 at 6:35
  • $\begingroup$ On a side note, I don't see how being finite union of closed sets is contained in the definition (I mean, it is apparent that it is union of one closed set, and thus of finitely many closed sets, but I would guess that's not what you're referring to) $\endgroup$ Mar 12 '19 at 6:38
  • $\begingroup$ each interval is a finite union of closed intervals and hence closed @SaucyO'Path $\endgroup$
    – Secretly
    Mar 12 '19 at 6:42
  • $\begingroup$ @hopefully Ah, ok. I was misreading. $\endgroup$ Mar 12 '19 at 6:47
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Let $C_n=\frac{C_{n-1}}{3}\cup \left(\frac{2}{3}+\frac{C_{n-1}}{3}\right)$ with $C_0=[0,1]$.

  • $C_n$ is closed

  • $C_n$ is contained in $[0,1]$

  • Every interval in $C_n$ is at maximum $(\frac{2}{3})^n$ in length

Cantor set is defined as:

$C:=\cap_{n=0}^{\infty}C_n$

By definition it is clear that

  • it is bounded (is contained in $[0,1]$)

  • it is closed (is the infinite intersection of closed sets)

  • Every neighborhood of a point in $C$ must contain another point of $C$at least: you can prove it using the third property of $C_n$

Note that this property is noteworthy, since the set is totally disconnected

Rigorous proof: By simple computations, you can note that the Cantor set consists of all the real numbers of the unit interval that do not require 1 in their ternary expansion. Thus, for every $c\in C=0.c_1c_2\dots$, there exists, for all $n$, a number $c_n \in C$ such that $|c-c_n|<3^{-n}$, constructed in this way: truncate $c$ to the $n+1$-th digit, and substitute it with $2$ if it is $0$ and viceversa. The number so constructed is still in $C$, and has a distance from $c$ of $\frac{2}{3}\cdot 3^{-n}$.

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  • $\begingroup$ could you please show me rigorously how to show that every point in the Cantor set is a cluster point? $\endgroup$
    – Secretly
    Mar 12 '19 at 7:01
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    $\begingroup$ @hopefully I inserted a proof, let me know if it is clear $\endgroup$
    – Caffeine
    Mar 12 '19 at 9:49

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