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Problem. Let $G = \{(x,y): x \ne y\}$. Prove $G$ is an open subset of $\mathbb R^2$.

What I am thinking: If I could show that $\mathbb R^2 \setminus G = \{(x,y): x = y\}$ is a closed set, then its complement $G$ is open. I might be totally off. Any suggestions?

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    $\begingroup$ How did you define an open/closed set? $\endgroup$ – Listing Nov 24 '13 at 20:38
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Let the function $f\colon \mathbb R^2\rightarrow \mathbb R,$ $(x,y)\mapsto x-y$ then $f$ is continuous since it's a polynomial on $x$ and $y$. We have $$G=f^{-1}(\mathbb R^*)$$ Do you know what theorem we should use to conclude?

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    $\begingroup$ Would the theorem be if f is continuous then the inverse image under f of any open set is open? I am only confused on what is meant by R*? $\endgroup$ – MDW Nov 24 '13 at 20:58
  • $\begingroup$ Yes this is the needed theorem and $\mathbb R^*$ is $\mathbb R\setminus \{0\}$. $\endgroup$ – user63181 Nov 24 '13 at 21:00
  • $\begingroup$ @Sami: Why can't we just show $\Bbb{R}^2 - G$ is open? $\endgroup$ – Don Larynx Nov 24 '13 at 21:03
  • $\begingroup$ Do you mean show that $\mathbb R^2-G$ is closed? Yes it's possible and I want avoid this since I don't know his definition of closed set @DonLarynx $\endgroup$ – user63181 Nov 24 '13 at 21:07
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    $\begingroup$ @SamiBenRomdhane: It is open if GOD wants. :D $\endgroup$ – mrs Nov 25 '13 at 2:40
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Just partition $G$ into two disjoint sets: Let $G_1 = \{(x,y) : x > y \}$ and $G_2 = \{(x,y) : x < y \}$, then clearly $G = G_1 \cup G_2$.

Now, WLOG, let's show that $G_1$ is open. Let $\textbf{x} = (x_1,y_1) \in G_1$, and $B_\textbf{x}(r)$ be the ball around $\textbf{x}$ of radius $r$. In order for $G_1$ to be open, we need to find an $r$ such that $B_\textbf{x}(r) \subset G_1$. But, just choose $r$ to be the shortest distance between $x$ and the line $y=x$. Informally, we know the shortest distance between a point to a line is just the intersection of the normal line with the point $\textbf{x}$, and the line y = x (in our case). The normal line will be $y = -x + x_1 + y_1$ (by a quick computation), and if we intersect this line with $y=x$, we get $x = \frac{x_1 + y_1}{2}$, so we can compute the distance to be $\frac{x_1 - y_1}{\sqrt 2}$. choose $r$ to be this, and we have shown $G_1$ will be open.

Do the same for $G_2$, then $G$ is the union of 2 open sets, and we are done.

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The definition of an open set : A subset $U$ of a metric space $(M,d)$ is called \textit{open} if, given any point $x$ in $U$, there exists a real number $ε > 04$ such that, given any point $y$ in $M$ with $d(x, y) < ε$, $y$ also belongs to $U$.

In the case of a line in the plane given by the equation $ax + by + c= 0$, where $a, b$ and $c$ are real constants with $a$ and $b$ not both zero, the distance from the line to a point $(x_0,y_0)$ is $$ \frac{|ax_0 + by_0 +c|}{\sqrt{a^2 + b^2}}.$$ Now, we are interested in one specific line, namely $y = x$, i.e. $x - y = 0$.

The distance of any point $(x_0, y_0)$ in $G$ to the line $ x -y = 0$ is therefore $$r :=\frac{|x_0 -y_0|}{\sqrt{2}}.$$

Now, consider the open ball centered in $(x_0,y_0)$ and of radius $r$. Do every element of this ball belong to $G$ ?

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Suggestion (you were looking for):

Use the theorem that says that the graph of a real continuous function is closed.

$A=\{(x,y)\in \mathbb{R}^2 : x=y\}$ is the graph of the continuous function $f:\mathbb{R}\rightarrow \mathbb{R}$. Where $$f(x)=x$$.

Hence A is closed. Therefore, its complement is open.


Addendum:

The theorem does not hold in general but it certainly does for Hausdorff spaces. All metric spaces ($\mathbb{R}$ is a metric space) are Hausdorff.

Another theorem that could be applied directly to solve this problem is:

Theorem: A topological space $X$ is Hausdorff iff its diagonal is closed.

The diagonal is defined as $$\{(x,x):x\in X\}$$

Note: The diagonal is closed in the product topology on $X\times X$.

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HINT: $\Bbb{R}^2 - G$ results in a line. Can this complete $G$?

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