2
$\begingroup$

I want to proof that the space $\{0,1\}^{\mathbb N}$ of infinite binary sequences with the product topology is totally disconnected. I know that this space has a basis consisting of clopen sets and is $T_2$, so it follows already that it is totally disconnected. But I tried to proof it directly using the definition.

Def: A topological space $X$ is totally disconnected iff only the singletons are connected, i.e. if for every set $Y$ with more than two elements there exists two nonempty separated sets $X_1, X_2$ such that $Y = X_1 \cup X_2$.

Proof: Let $Y \ne X$ be a set with more than one element. Then if $Y$ is finite, select some $y \in Y$, and because for finite sets $M$ it holds that $cl(M) = M$ it follows that $Y = \{y\} \cup Y\setminus\{y\}$ is a separation. Otherwise $Y$ is infinite if we can select a point $y$ which is not a limit point of $Y$ and we can write again $Y = \{y\} \cup Y\setminus\{y\}$.

I am not sure if it is always possible to select a point which is not a limit point of $Y$, but I had sets like $Y = \{ ab^{\mathbb N}, aab^{\mathbb N},aaab^{\mathbb N},\ldots \}$ in mind, which has limit point $a^{\mathbb N}$. Is there a way to proceed along the lines of this proof and construct for every $Y$ such a partition into separated sets?

$\endgroup$
  • 1
    $\begingroup$ "I am not sure if it is always possible to select a point which is not a limit point of $Y$" It isn't. The space has perfect proper subsets. $\endgroup$ – Daniel Fischer Sep 20 '13 at 13:35
  • 2
    $\begingroup$ I would argue: Let $x \in X$, and $C$ the connected component of $X$. Since the projection $\pi_k$ to the $k$-th coordinate is continuous, we have $\pi_k(C) \subset \pi_k(\{x\})$. Since that holds for all $k$, $C \subset \prod \pi_k^{-1}(x_k) = \{x\}$. $\endgroup$ – Daniel Fischer Sep 20 '13 at 13:38
5
$\begingroup$

In $\{0,1\}^\Bbb{N}$ let $A$ be a subset with at least two elements $(a_n)_n$ and $(b_n)_n.$ There is an index $k$ where $a_k\ne b_k.$ But then $p_k^{-1}(0)$ and $p_k^{-1}(1)$ are disjoint open sets ($p_k$ is the projection onto the $k$-th coordinate), one containing $(a_n)_n$ and the other one containing $(b_n)_n$, and both cover $A.$

$\endgroup$

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.