# Show that X is compact

#### Alexmahone

##### Active member
Consider the two-element set $\{0,\ 1\}$ equipped with the discrete topology, and form the countably infinite product

$\displaystyle X:=\{0,\ 1\}^\omega=\prod\limits_{n\in\mathbb{Z_+}}\{0,\ 1\}$

So $X$ consists of the infinite sequences $\displaystyle(x_n)_{n\in\mathbb{Z_+}}$, where for each $k\in\mathbb{Z}_+$, the $k$th term $x_k$ is either $0$ or $1$. Equip $X$ with the product topology.

Show that $X$ is compact (you may not use Tychonoff’s theorem).

#### GJA

##### Well-known member
MHB Math Scholar
Hi Alexmahone,

One way to proceed is to take an arbitrary open cover $\{U_{\alpha}\}_{\alpha\in\mathcal{I}}$ and prove that it must admit a finite subcover by using the definition of the product topology. Try giving that a shot and see what you can come up with.

#### Janssens

##### Well-known member
Yes, first do what post #2 says.

You can make it a bit simpler for yourself by restricting to basic-open covers: Show that every cover consisting of basic-open sets has a finite subcover.

(Alternatively, if your book defines the product topology in terms of a subbasis, you can consider using Alexander's theorem, but I don't know whether that theorem can be proven without Zorn's lemma. Since "Zorn" is equivalent to "Tychonoff", this route may therefore be implicitly forbidden and, in any case, it uses more force than needed.)

#### Alexmahone

##### Active member
Let $\mathcal{A}=\{A_\alpha\}$ be an open cover of $X$.

$\implies A_\alpha$'s are open sets of $X$ and $\bigcup\limits_\alpha A_\alpha=X$.

The problem is that it's not easy to describe the $A_\alpha$'s. The basis elements of $X$ can be easily described: they are infinite cartesian products whose components are subsets of $\{0, 1\}$.

And I haven't learnt that it is enough to restrict myself to basic-open covers.

#### Janssens

##### Well-known member
The problem is that it's not easy to describe the $A_\alpha$'s. The basis elements of $X$ can be easily described: they are infinite cartesian products whose components are subsets of $\{0, 1\}$.
A basic-open element is a finite intersection of sets of the form $\pi_k^{-1}(O_k)$ where $O_k$ is any subset of $\{0,1\}$ and $\pi_k$ is the $k$th coordinate projection.

And I haven't learnt that it is enough to restrict myself to basic-open covers.
It is probably not really necessary, but this is not difficult to prove by itself, as a lemma for your problem.

#### GJA

##### Well-known member
MHB Math Scholar
The basis elements of $X$ can be easily described: they are infinite cartesian products whose components are subsets of $\{0, 1\}$.
Since you know what the basis elements are, take one open set from your cover and use what you know about basis elements. You will need to reduce the problem down to something you know about (i.e., the basis elements).

#### caffeinemachine

##### Well-known member
MHB Math Scholar
Consider the two-element set $\{0,\ 1\}$ equipped with the discrete topology, and form the countably infinite product

$\displaystyle X:=\{0,\ 1\}^\omega=\prod\limits_{n\in\mathbb{Z_+}}\{0,\ 1\}$

Each element of $X$ can be thought of as a point in the Cantor set. Here is how I like to see it.

So $X$ consists of the infinite sequences $\displaystyle(x_n)_{n\in\mathbb{Z_+}}$, where for each $k\in\mathbb{Z}_+$, the $k$th term $x_k$ is either $0$ or $1$. Equip $X$ with the product topology.

Show that $X$ is compact (you may not use Tychonoff’s theorem).
Consider the infinite binary rooted tree and call it $T$. (It has a root vertex which has two children, each of these two children has two children and so on. Every child has exactly one parent.)

Every element of $X$ can be thought of as an infinite path starting at the root. Thus there is a natural bijection $f:X\to P(T)$, where $P(T)$ is the set of all the infinite paths in $T$ starting at the root. This bijection gives us a topology on $P(T)$. Also, $P(T)$ itself can naturally be identified with the Cantor set.

Not trying to write all details, let me just add one useful piece of information which may help proving that $P(T)$ is homemorphic to the Cantor set. For a vertex $v\in T$, define $B_v\subseteq P(T)$ as $B_v=\{\gamma\in P(T):\ \gamma \text{ passes through } v\}$.
Note that, by definition of the product topology, $\{B_v\}_{v\in T}$ forms a basis for the topology on $P(T)$.