How to show the diagonal in the extended plane is closed?

[psie]

TL;DR

Consider $\{(x,x):x\in\mathbb R\}$. I've managed to show this set is closed in $\mathbb R^2$ by showing its complement is open. Now using what I've already showed, how can I argue that $\{(x,x):x\in\overline{\mathbb R}\}$ is closed in $\overline{\mathbb R}^2$?

Let $(f_n)$ be a sequence of measurable functions from $E$ into $\mathbb R$. I'm reading a proof of the fact that the set $A$ of all $x\in E$ for which $f_n(x)$ converges in $\mathbb R$ as $n\to\infty$ is measurable. The proof goes like this (I'm paraphrasing):

Define $G:E\to\overline{\mathbb R}^2$ by $G(x)=(\liminf f_n(x),\limsup f_n(x))$. Then $G$ is measurable and if $D=\{(x,x):x\in\mathbb R\}$, then $$A=\{x\in E:-\infty<\liminf f_n(x)=\limsup f_n(x)<\infty\}=G^{-1}(D).$$So the measurability of $A$ follows since $D$ is measurable subset of $\overline{\mathbb R}^2$ (note that $\{(x,x):x\in\overline{\mathbb R}\}$ is measurable as a closed subset of $\overline{\mathbb R}^2$).

Why is $\{(x,x):x\in\overline{\mathbb R}\}$ closed in $\overline{\mathbb R}^2$? As I said, I can show the diagonal with $x\in\mathbb R$ is closed in $\mathbb R^2$, but if $x\in \overline{\mathbb R}$ and the space is $\overline{\mathbb R}^2$, how does one proceed?

 

[WWGD]

TL;DR Summary: Consider $\{(x,x):x\in\mathbb R\}$. I've managed to show this set is closed in $\mathbb R^2$ by showing its complement is open. Now using what I've already showed, how can I argue that $\{(x,x):x\in\overline{\mathbb R}\}$ is closed in $\overline{\mathbb R}^2$?

Let $(f_n)$ be a sequence of measurable functions from $E$ into $\mathbb R$. I'm reading a proof of the fact that the set $A$ of all $x\in E$ for which $f_n(x)$ converges in $\mathbb R$ as $n\to\infty$ is measurable. The proof goes like this (I'm paraphrasing):



Why is $\{(x,x):x\in\overline{\mathbb R}\}$ closed in $\overline{\mathbb R}^2$? As I said, I can show the diagonal with $x\in\mathbb R$ is closed in $\mathbb R^2$, but if $x\in \overline{\mathbb R}$ and the space is $\overline{\mathbb R}^2$, how does one proceed?

Is $\overline{\mathbb R}$ the 1-pt compactification? What metric do you use in $\overline{\mathbb R}^2$; the product metric? Edit: Is $\overline{\mathbb R}^2=\overline {\mathbb R} \times \overline {\mathbb R}$?

Edit 2: Isn't the compactification map an embedding , and, if so, it sends closed sets to closed sets.

 

[martinbn]

Why does you proof for $\mathbb R$ doesn't work for $\overline{\mathbb R}$? In any case for a topological space $X$, the diagonal is closed in $X\times X$ (with the product topology) if and only if $X$ is Hausdorff. This is easy to check. And your $\overline{\mathbb R}$, I suppose is Hausdorff.

 

[psie]

Good questions.

Is $\overline{\mathbb R}$ the 1-pt compactification? What metric do you use in $\overline{\mathbb R}^2$; the product metric? Edit: Is $\overline{\mathbb R}^2=\overline {\mathbb R} \times \overline {\mathbb R}$?

Yes, $\overline{\mathbb R}$ is $\{-\infty\}\cup \{\infty\}\cup \mathbb R$. The ad hoc metric is I think $d(x,y)=|\arctan x-\arctan y|$, so it is homeomorphic to $[-\pi/2,\pi/2]$. Yes, $\overline{\mathbb R}^2=\overline {\mathbb R} \times \overline {\mathbb R}$.

Edit 2: Isn't the compactification map an embedding , and, if so, it sends closed sets to closed sets.

I have only very rudimentary tools at my disposal (metric spaces and the fact that $\overline {\mathbb R}$ is metrizable), so I don't understand your Edit 2.

Why does you proof for $\mathbb R$ doesn't work for $\overline{\mathbb R}$?

It doesn't work because the proof uses the Euclidean norm on $\mathbb R^2$. We can't equip $\overline {\mathbb R}^2$ with the Euclidean norm.

 

[martinbn]

It doesn't work because the proof uses the Euclidean norm on $\mathbb R^2$. We can't equip $\overline {\mathbb R}^2$ with the Euclidean norm.

You take something off the diagonal $(x,y)$, this means $x\not = y$. Then you take to open sets $U$ and $V$ that don't intersect and contain $x$ and $y$ respectively. Then $U\times V$ is an open set containing $(x,y)$ and not meeting the diagonal.