# Product Topology

• Feb 9th 2010, 04:51 PM
Drexel28
Product Topology
Hello everyone. I have the following problem.

Quote:

Let $\displaystyle X_j=[0,1]$ with the usual topology for every $\displaystyle j\in[0,1]$ and let $\displaystyle X=\prod_{j\in[0,1]}X_j$. Prove that $\displaystyle X$ is not second countable.
The quote says to consider that if $\displaystyle X$ were second countable that every open base must have a countable subset which is also an open base.

But, I don't understand why this doesn't work.

Clearly we have that $\displaystyle G_j=\prod_{k\in[0,1]}E_k$ where $\displaystyle E_k=\begin{cases} (0,1) & \mbox{if} \quad k=j \\ X_k & \mbox{if} \quad k\ne j\end{cases}$ is open in $\displaystyle X$ (in fact it's a subbasic open set). So, we have that $\displaystyle O=\bigcup_{j\in[0,1]}G_j$ is an open set in $\displaystyle X$. If $\displaystyle X$ were second countable we would have, by Lindelof's theorem, that $\displaystyle O$ must have a countable subcover. But, that isn't the case here. Right?
• Feb 10th 2010, 03:22 PM
Drexel28
Quote:

Originally Posted by Drexel28
Hello everyone. I have the following problem.

The quote says to consider that if $\displaystyle X$ were second countable that every open base must have a countable subset which is also an open base.

But, I don't understand why this doesn't work.

Clearly we have that $\displaystyle G_j=\prod_{k\in[0,1]}E_k$ where $\displaystyle E_k=\begin{cases} (0,1) & \mbox{if} \quad k=j \\ X_k & \mbox{if} \quad k\ne j\end{cases}$ is open in $\displaystyle X$ (in fact it's a subbasic open set). So, we have that $\displaystyle O=\bigcup_{j\in[0,1]}G_j$ is an open set in $\displaystyle X$. If $\displaystyle X$ were second countable we would have, by Lindelof's theorem, that $\displaystyle O$ must have a countable subcover. But, that isn't the case here. Right?

Just as a follow-up. I posted this on the AOPS website and got an affirmative response.

Clearly $\displaystyle O$ is an open set engineered to have an uncountable open cover with no proper subcover. The fact that it cannot satisfy Lindelof's theorem, and thus cannot be second countable follows.
• Feb 10th 2010, 03:51 PM
mabruka
Right, i agree it can not have a countable subcover.

Suppose there is, $\displaystyle O \subset \bigcup_{j=i_1,i_2,...}G_j$ for some numbers $\displaystyle \{i_1,i_2,...\} \subset [0,1]$

Now let $\displaystyle x \in X$ be such an element that $\displaystyle \pi_j (x) = 1$ for every $\displaystyle j \in \{i_1,i_2,...\}$ where $\displaystyle \pi_j$ is the projection on the jth coordenate.

Clearly $\displaystyle x \not \in\bigcup_{j=i_1,i_2,...}G_j$ but $\displaystyle x \in \bigcup_{j\in[0,1]}G_j = O$

What do you think?
• Feb 10th 2010, 04:01 PM
Drexel28
Quote:

Originally Posted by mabruka
Right, i agree it can not have a countable subcover.

Suppose there is, $\displaystyle O \subset \bigcup_{j=i_1,i_2,...}G_j$ for some numbers $\displaystyle \{i_1,i_2,...\} \subset [0,1]$

Now let $\displaystyle x \in X$ be such an element that $\displaystyle \pi_j (x) = 1$ for every $\displaystyle j \in \{i_1,i_2,...\}$ where $\displaystyle \pi_j$ is the projection on the jth coordenate.

Clearly $\displaystyle x \not \in\bigcup_{j=i_1,i_2,...}G_j$ but $\displaystyle x \in \bigcup_{j\in[0,1]}G_j = O$

What do you think?

Yes. I agree. Here is what I said. Let $\displaystyle \bold{x}_\ell$ be such that $\displaystyle \pi_\ell\left(\bold{x}_\ell\right)=\frac{1}{2}$ and $\displaystyle \pi_{k}\left(\bold{x}_j\right)=1,\text{ }k\ne \ell$. Clearly we have that $\displaystyle \bold{x}_\ell$ is only in $\displaystyle G_\ell$ and so removing it from $\displaystyle \left\{G_j\right\}_{j\in[0,1]}$ will result in a cover that no longer covers $\displaystyle O$. Since $\displaystyle \ell$ was arbitrary it follows that the removal of any set in $\displaystyle \left\{G_j\right\}_{j\in[0,1]}$ will result in a non-cover. Thus, $\displaystyle \left\{G_j\right\}_{j\in[0,1]}$ has no proper subcover and thus clearly no countable subcover.