Welcome to our community

Be a part of something great, join today!

Problem of the Week #53 - June 3rd, 2013

Status
Not open for further replies.
  • Thread starter
  • Moderator
  • #1

Chris L T521

Well-known member
Staff member
Jan 26, 2012
995
Here's this week's problem.

-----

Problem: Let $J$ be an uncountable index set. Prove that $\mathbb{R}^J$ in the product topology is not metrizable.

-----

Remember to read the POTW submission guidelines to find out how to submit your answers!
 
  • Thread starter
  • Moderator
  • #2

Chris L T521

Well-known member
Staff member
Jan 26, 2012
995
This week's problem was correctly answered by Opalg. You can find his solution below.

Claim 1. If $X$ is a metric space and $E\subset X$, then every element $x$ in the closure of $E$ is the limit of a sequence in $E$.

Proof. For $n=1,2,\ldots$, the ball of radius $1/n$ centred at $x$ has nonempty intersection with $E$. Let $x_n$ be an element in that intersection. Then $x_n\to x.$


Now define a subset $E\subset \mathbb{R}^J$ as follows. Denote by $\mathcal{F}$ the set of all finite subsets of $J$. For each $F\in\mathcal{F}$ define $x_F\in \mathbb{R}^J$ by $(x_F)_j = \begin{cases}0& (j\in F), \\ 1& (j\notin F). \end{cases}$
So $E$ is the set of all elements of $\mathbb{R}^J$ whose coordinates are all $1$ except for a finite number that can be $0$. Denote by $\mathbf{0}$ the element of $\mathbb{R}^J$ whose coordinates are all zero.


Claim 2. The closure of $E$ contains $\mathbf{0}$.

Proof. It will be sufficient to show that every basic neighbourhood of $\mathbf{0}$ (for the product topology) has nonempty intersection with $E$. A basic neighbourhood $U$ of $\mathbf{0}$ has the following form. First, take $F\in \mathcal{F}$. Next, for each $j\in F$, let $U_j$ be a neighbourhood of $0\in\mathbb{R}$. Then $U = \{x = (x_j)_{j\in J} \in \mathbb{R}^J : x_j\in U_j\;(\forall j\in F)\}.$ Given such a neighbourhood, clearly, $x_F \in U\cap E.$ So $U\cap E$ is nonempty, as required.


Claim 3. There is no sequence in $E$ that converges to $\mathbf{0}$.

Proof. Let $\bigl(x_{F_n}\bigr)_{n\geqslant1}$ be a sequence in $E$. Then \(\displaystyle F = \bigcup_{n\geqslant1}F_n\) is a countable union of finite subsets of $J$ and is therefore countable. Since $J$ is uncountable, there exists an element of $J_0\in J$ that is not in $F$ and therefore not in any of the sets $F_n.$ The set $\{x\in \mathbb{R}^J : x_{j_0} <1\}$ is a neighbourhood of $\mathbf{0}$ that does not contain any of the points $x_{F_n}$ (because they all have $j_0$-coordinate equal to $1$). Therefore the sequence $(x_{F_n})$ cannot converge to $\mathbf{0}$ in the product topology.


The three Claims together show that the product topology on $\mathbb{R}^J$ is not metrisable.
 
Status
Not open for further replies.