# Show the existence of a locally compact, perfect set in $[0,1]$ which has the following property.

Question: Show that there exists a Locally Compact, Perfect set in $$[0,1]$$ such that ternary expansion of each of its points consists of $$0$$ and $$1$$ only.

I know that the Cantor Set has almost all the above properties except the last one.

$$1.$$ Cantor set is compact and hence locally compact.

$$2.$$ Cantor set is a perfect set.

$$3.$$ Cantor set consists of members whose ternary expansion consists of $$0$$ and $$2$$ only. But here we required $$0$$ and $$1$$ only. How should I proceed? Thanks in advance!

• First off, note that the subset will automatically be bounded, and perfect implies closed, so locally compact is superfluous, as such a set will be compact. I would proceed by choosing one set with only $0$ and $1$ in its ternary expansion, and then checking if it hit the mark. For instance, take the set of all numbers in $[0,1]$ that have only $0$ and $1$ in their ternary expansion. Is it perfect? If it is, then cool, you're done. If it isn't, is it because it isn't closed, or because it has isolated points? (Maybe both?) Can it be fixed in some simple way? Jun 21 '19 at 10:40
• A hint: look for such sets using the 9-ary expansion.
– rts
Jun 22 '19 at 2:59
• Both commenters thanks for the help. Jun 22 '19 at 7:08

Apply $$f:[0,1] \to [0,1]$$ defined by $$f(x)=\frac{x}{2}$$ to the middle third Cantor set $$C$$. $$f[C]$$ is clearly homeomorphic to $$C$$ and fits your requirements:
If $$x\in C$$ is written as $$x=\sum_{n \ge 1} \frac{a_n}{3^n}$$ with all $$a_n \in \{0,2\}$$, then $$f(x) = \sum_{n \ge 1} \frac{b_n}{3^n}$$ with $$b_n = \frac{a_n}{2} \in \{0,1\}$$.
• Sir, what is middle third Cantor set? Did you mean the Cantor set in $[\frac{2}{3},1]$? Jun 22 '19 at 7:10
• @SujitBhattacharyya It's what you call "Cantor set" (in $[0,1]$ constructed, by throwing out the "middle thirds" of all closed intervals in successive stages, hence the name). Topologists call a Cantor set any space/set homeomorphic to it, like $\{0,1\}^{\Bbb N}$ etc. So I mean the Cantor set you are talking about, which has ternary digits $0$ and $2$ and these are halved by $f$ and hence we get ternary digits $0$ and $1$ instead. Jun 22 '19 at 7:30