Why the Cantor set have $\mathfrak{c}$ subsets homeomorphic to $\mathbb{Q}$?

Observe that $${^{2}({^{\omega}2})}\thickapprox{^{\omega}({^{\omega}2})}\thickapprox{^{\omega}2}$$. Consequently, each Cantor set contains a family of $$\mathfrak{c}$$ pairwise disjoint Cantor sets. If the Cantor set $$2^{\omega}$$ contains at least one homeomorphic subset to $$\mathbb{Q}$$, then $$2^{\omega}$$ has at least $$\mathfrak{c}$$ homeomorphic subsets to $$\mathbb{Q}$$.

Now, why is there at most $$\mathfrak{c}$$ homeomorphic subsets to $$\mathbb{Q}$$?

And how can I prove that $$2^{\omega}$$ has at least one homeomorphic subset to $$\mathbb{Q}$$?

• Another way to find a subset homeomorphic to $\Bbb Q$ is to note that, as an ordered set, the Cantor set is the Dedekind completion of a set of order type $1+2\cdot\Bbb Q+1$, the set of endpoints of removed intervals Commented Jul 11, 2019 at 9:21

The first question is just a consequence of cardinal arithmetic: the number of subsets of size $$\kappa$$ of a set of size $$\lambda$$ is bounded above by the set of functions from $$\kappa$$ to $$\lambda$$, which is just $$\lambda^\kappa$$ - now observe that $$(2^{\aleph_0})^{\aleph_0}=2^{\aleph_0}$$.
In the other direction, note that $$(1)$$ every countable dense subset of $$\mathbb{R}$$ is homeomorphic to $$\mathbb{Q}$$ (indeed, any two countable dense linear orders without endpoints are isomorphic), and $$(2)$$ the Cantor set is quite similar to $$\mathbb{R}$$ - specifically, $$\mathbb{R}$$ is the image of the Cantor set under a continuous map which is injective except on a countable set. This suggests that to find something $$\mathbb{Q}$$-like in the Cantor set we should start by considering arbitrary countable dense subsets of the Cantor set; can you find one, and determine whether it is homeomorphic to $$\mathbb{Q}$$?
• Thank you! Your comment on the second question reminded me that any zero-dimensional space (for example $\mathbb{Q}$) can be imbedded in the Cantor's set (Corollary 1.5.7 - The infinite-dimensional topology of function spaces). Now I have everything. Commented Jul 10, 2019 at 20:00
• @FernandoMauricioRiveraVega Unless I'm missing something, that's not right as stated - consider a discrete space of cardinality $>2^{\aleph_0}$. Commented Jul 10, 2019 at 20:21