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?

  • 1
    $\begingroup$ It's contained in the unit interval. $\endgroup$ – Lord Shark the Unknown Mar 12 at 6:34
  • $\begingroup$ @LordSharktheUnknown thank you .... so I will edit my question to another one $\endgroup$ – hopefully Mar 12 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) $\endgroup$ – Saucy O'Path Mar 12 at 6:38
  • $\begingroup$ each interval is a finite union of closed intervals and hence closed @SaucyO'Path $\endgroup$ – hopefully Mar 12 at 6:42
  • $\begingroup$ @hopefully Ah, ok. I was misreading. $\endgroup$ – Saucy O'Path Mar 12 at 6:47

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:


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 $\frac{2}{3}\cdot 3^{-n}$.

  • $\begingroup$ could you please show me rigorously how to show that every point in the Cantor set is a cluster point? $\endgroup$ – hopefully Mar 12 at 7:01
  • 1
    $\begingroup$ @hopefully I inserted a proof, let me know if it is clear $\endgroup$ – Gabriele Cassese Mar 12 at 9:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.