How to show $B=\{(x,y):x=y\}$ is closed?

Question

Is $B=\{(x,y):x=y\}$ in $\mathbb{R}^2$ closed? open? bounded?

So I am almost sure that $B$ is closed and not bounded, because it is built just form boundary points.

I started with looking at $B^{C}=\{(x,y):x\neq y\}$, now it is left to show that for every $a\in B$ there is $r>0$ such that $B(a,r)\subset B^{C}$

How should continue? I have tried to look at

$||a-p||$$=||(x_{0},y_{0})-(x,y)||=\sqrt{(x_{0}-x)^2+(y_{0}-y)^2}$ and to prove that it smaller then an expression (=r)

Second thought: $B^{C}$ is 2 set divided by a line, can it be any useful?

Answer 1

The simplest way if probably to look at the continuous function $f\colon \mathbb{R}^2\to \mathbb{R}$ defined by $f(x,y)=x-y$. It is continuous, and $\{0\}$ is a closed set in $\mathbb{R}$, so $f^{-1}(\{0\})$ is closed in $\mathbb{R}^2$.

As for bounded... Choose any sequence $(x_n)_n$ going to $\infty$, and consider the sequence $(x_n,x_n)_n$ in $B$.

Answer 2

Just in general when you encounter something like this later: $X$ is Hausdorff $\iff$ the diagonal of $X$, denoted $\Delta_X$ is closed.

Answer 3

To show the complement is open, which is how you started, choose any $x\neq y$ and note that there are open intervals $I_x$ and $I_y$ such that $I_x\cap I_y=\emptyset .$ Now consider the open square $I_x\times I_y\subset \mathbb R^2$. If $a\in I_x$ then $a\notin I_y$ so that $I_x\times I_y$ contains no point $(a,a)$. i.e. $I_x\times I_y\subset B^c.$

edit: if you are required to use open balls, without loss of generality assume that $\vert I_x\vert =\vert I_y\vert =l$ and consider any ball $\mathcal B$ of radius $<l/2$, whose center is located in the center of the square.Then $\mathcal B$ is disjoint from the diagonal.

Answer 4

The fact that $B$ is closed can be deduced from the following more general (and more peculiar) result:

Proposition. A topological space $(X,\tau)$ is Hausdorff if and only if its diagonal $\Delta_X:=\lbrace (x,x):x\in X\rbrace$ is closed in $X\times X$ (with the product topology).

Proof. Let us write $\mathscr{B}_\Pi:=\lbrace U\times V: U,V\in\tau\rbrace$ to denote the base for the product topology on $X\times X$. Moreover, we write $\mathscr{U}_x:=\lbrace U\in \tau: x\in U\rbrace$ to denote the set of all open neighbourhoods of $x$ in $X$.

Suppose that $\Delta_X$ is closed. Then $X\times X\setminus \Delta_X$ is open. Suppose that $x$ and $y$ are two distinct points in $X$. Then clearly $(x,y)$ is not an element of the diagonal, and thus $(x,y)\in X\times X\setminus \Delta_X$. Since this set is open, we may find a base element $U\times V$ in the base $\mathscr{B}_\Pi$ such that $(x,y)\in U\times V$. These two open sets $U,V\in \tau$ must be disjoint, for otherwise we could find $z\in U\cap V$ which would imply $(z,z)\in U\times V$, so that $U\times V \nsubseteq X\times X\setminus \Delta_X$. Thus $(X,\tau)$ is Hausdorff.

Now suppose that $\Delta_X$ is not closed. This means that $X\times X\setminus \Delta_X$ is not open. This means that we can find $(x,y)\in X\times X\setminus \Delta_X$ so that for all open neighbourhoods $W$ of $(x,y)$ there exists a point $(p,q)\in W$ such that $(p,q)\notin X\times X\setminus \Delta_X$. The fact that $(x,y)\in X\times X\setminus \Delta_X$ means that $x$ and $y$ are distinct. The fact that $(p,q)\notin X\times X\setminus \Delta_X$ means that $p=q$. Since in particular base elements in $\mathscr{B}_\Pi$ are open, we may take $W=U\times V$ for some $U\in \mathscr{U}_x$ and $V\in \mathscr{U}_y$. Thus, the above statement is equivalent to saying that we can find two distinct points $x,y\in X$ so that for every open neighbourhoods $U\in \mathscr{U}_x$ and $V\in\mathscr{U}_y$ there exists a point $(p,p)\in U\times V$. But this means simply that $p\in U\cap V$, and thus $U$ and $V$ are not disjoint. Hence distinct points can not be separated by disjoint open sets, meaning that $(X,\tau)$ is not Hausdorff. The result now follows by contraposition. Q.E.D.

The answer to your question now follows by taking $X=\mathbb{R}$ with the Euclidean topology, which is of course Hausdorff, and noting that $\Delta_\mathbb{R}=B$.

(Edit. See also Faraad Armwood's answer.)

Answer 5

A map $f: \mathbb R \to \mathbb R$ is said to be a closed function if its graph $$G(f)=\{(x,y) \mid x \in \mathrm{Dom}(f)\,\,, y= f(x)\} \subseteq \mathbb R^2$$ is closed in the product topology.

Lemma: an everywhere defined continuous map is closed.

proof: Note the following criteria: $G(f)\, \mathrm{is \,closed} \iff $ if $x_n \to x$ and $f(x_n) \to y$, then $(x,y) \in G(T)$ and $f(x)=y$.

Suppose that $x_n \to x$ and $f(x_n) \to y$. We know that $x \in \mathrm{Dom(f)}$, since it is everywhere defined, and $f(x)=f(\lim_{n \to \infty} x_n)=\lim_{n \to \infty}f(x_n)=y,$ since continuous implies sequentially continuous.

theorem: $f:\mathbb R \to \mathbb R$ where $f(x)=x$ is a closed function.

proof: immediate from the previous lemma since the identity map is continuous.