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To prove that a set is Cantor, we have to prove that it's closed and bounded (compact), contains no intervals of positive length, and is perfect. The union of two Cantor sets would also be compact since union of closed bounded sets are still closed and bounded, but I'm having trouble proving whether it contains intervals of positive length or if it's perfect.

Would the union be a Cantor set? How would I go about proving/disproving it's perfect and contains no intervals of positive length?

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  • $\begingroup$ use de morgan's laws $\endgroup$
    – JMP
    Nov 9 '15 at 21:33
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Suppose $K_1,K_2$ are Cantor.

That $K_1 \cup K_2$ is perfect: Take a point $x$ in the union. Then $x$ is in one of the sets. Hence it's a limit point of that set. Hence it's a limit point of the union.

No intervals of positive length: Suppose $(a,b) \subset K_1\cup K_2.$ We can choose $x \in (a,b)$ such that $x \not \in K_1.$ Then for $r>0$ small enough, $(x-r,x+r)\cap K_1 = \emptyset$ and $(x-r,x+r) \subset (a,b).$ Hence $(x-r,x+r)\subset K_2,$ contradiction.

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  • $\begingroup$ why is $(x-r,x+r)$ not a subset of $K_2$? $\endgroup$
    – oper
    Nov 9 '15 at 21:58
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    $\begingroup$ $K_2$ contains no intervals of positive length. $\endgroup$
    – zhw.
    Nov 9 '15 at 21:59
  • $\begingroup$ I edited the answer slightly to make it clearer. $\endgroup$
    – zhw.
    Nov 9 '15 at 22:05
  • $\begingroup$ Can you explain the middle statement? Why is $(x-r, x+r)\in (a,b)$ and how does $(x-r, x+r)\cap K_1=0$ help us? $\endgroup$
    – oper
    Nov 9 '15 at 22:32
  • $\begingroup$ $(x-r,x+r)\subset (a,b)$ for small $r$ because $(a,b)$ is open. $(x-r,x+r)\cap K_1 = \emptyset$ tells us $(x-r,x+r)\subset K_2.$ $\endgroup$
    – zhw.
    Nov 10 '15 at 0:15
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Hint for contains no intervals part: Let $I$ be any positive-length interval. We want to show that $I$ is not a subset of the union (i.e. the union does not contain an arbitrarily chosen interval $I$). There is a subinterval $J$ of $I$ that contains no points of the first Cantor set (why?), and there is a subinterval $J'$ of $J$ that contains no points of the second Cantor set (why?). Since $J'$ contains no points from either the first Cantor set or the second Cantor set (why?), it follows that $J'$ contains no points from the union, and thus the union does not contain $I$ (because the union does not contain the points in $I$ that belong to $J'$.)

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Showing that the union of two Cantor sets is perfect is just a direct application of the definition of "perfect." The other claim is a bit harder.

Towards containing no intervals of positive length, here is a hint: Try to prove the following strengthening of "Cantor sets don't contain intervals of positive length":

Suppose $X$ is a Cantor set, and $I$ is an interval of positive length. Then there is a sub-interval of positive length, $J\subseteq I$, which is disjoint from $X$ - that is, $J\cap X=\emptyset$.

(Note that this isn't trivial: $\mathbb{Q}$ doesn't contain any interval of positive length, but there are also no intervals of positive length disjoint from $\mathbb{Q}$. You'll need to use some special property of Cantor sets . . .)

If you can prove this, then: suppose $I$ is an interval of positive length, and $C_0, C_1$ are Cantor sets. Pass to a $J_0\subseteq I$ disjoint from $C_0$, and then a $J_1\subseteq J_0$ disjoint from $C_1$, each of positive length. What can you say about $J_1\cap C_0\cup C_1$?

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