Let $\Box_{i\in I} X_i$ denote the box product of the spaces $X_i$. The box product $\Box_{n\in\omega}\mathbb{R}$ is not connected, since the collection of bounded sequences is both open and closed.
Is $\Box_{n\in\omega}[0,1]$ connected?
$\begingroup$
$\endgroup$
1
asked Apr 7, 2018 at 12:05
-
2$\begingroup$ The set of sequences $(a_n)$ such that $(n a_n)$ is bounded seems to be open and closed, right? $\endgroup$– Mateusz KwaśnickiCommented Apr 7, 2018 at 12:48
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
3
It's not connected.
Let $u=(u_n)$ be a sequence. For $\ell\in [0,1]$, define $$V_{u,\ell}=\{(v_n):\forall n\in \omega:|v_n-\ell|<\max(2^{-n},2|u_n-\ell|)\}.$$ Then $u\in V_{u,\ell}$ and $V_{u,\ell}$ is open.
Moreover, if $\ell$ is a limit point of $u$ and $v\in V_{u,\ell}$, then $\ell$ is a limit point of $v$.
Let $L_\ell$ be the set of sequences in $[0, 1]$ converging to $\ell$, and $L_\ell^c$ its complement. Then on the one hand, $L_0=\bigcup_{u\in L_0}V_{u,0}$, which is thus open. Its complement $L_0^c$ is $\bigcup_{u\in L_0^c,\ell\in P(u,0)}V_{u,\ell}$, where $P(u,\ell)$ is the set of limit points $\neq\ell$ of $u$. So it's open as well. Thus $L_0$ is clopen. (Similarly $L_\ell$ is open for every $\ell$.
-
$\begingroup$ It seems that $M_\ell$ and $L_\ell$ have gotten slightly mixed up? $\endgroup$ Commented Apr 7, 2018 at 14:20
-
$\begingroup$ Another way to see that $M_0$ is closed is to note that it is closed in the uniform topology, which is weaker than the box topology. $\endgroup$ Commented Apr 7, 2018 at 14:21
-
$\begingroup$ Yep $M_\ell=L_\ell$ sorry. Feel free to edit. $\endgroup$– YCorCommented Apr 7, 2018 at 15:11