$O=\{(x,y)\in {R}^2: x

Question

can someone help me to show that $O=\{(x,y)\in {R}^2: x<y \}$ is an open Thanks. I have tried many things but i'm always stuck to a contradition for finding the radius

Answer 1

Let $f:\mathbb{R}^2 \to \mathbb{R}$ be the function defined by $f(x,y)=x-y$. Observe that $f$ is continuous and $f^{-1} [(-\infty,0)]=O$,l. Since $(-\infty,0)$ is open (in $\mathbb{R}$) then $O$ is open (in $\mathbb{R}^2$).

Answer 2

Let $p\in O $. Let $\epsilon \gt 0$ be the distance from $p$ to the line $y=x $. If $p=(x_0,y_0 ) $, that distance is $\epsilon = \sqrt { {(tx_0)}^2+{(ty_0)}^2} $, where $x_0-ty_0=y_0+tx_0$, or $t=\frac {x_0-y_0}{x_0+y_0} $. Then the ball $B (p,\frac \epsilon 2) $ of radius $\frac \epsilon 2$ around $p $ is contained in $O $, that is, $B (p,\frac \epsilon 2)\subset O $...

Answer 3

Let $(x_0,y_0)\in\mathbb{R}^2$ such that $x_0<y_0$. Let $r=\frac{y_0-x_0}{\sqrt2}$. Note that $r$ is the distance from $(x_0,y_0)$ to the line $y=x$. If $(x,y)\in\mathbb{R}^2$ is such that$$\bigl\|(x,y)-(x_0,y_0)\bigr\|<r=\frac{y_0-x_0}{\sqrt2},$$I will prove that $x<y$. Suppose otherwise, that is, suppose that $x\geqslant y$. Consider the points of $\mathbb{R}^2$ of the form$$\tag{1}t(x,y)+(1-t)(x_0,y_0),$$with $t\in[0,1]$; this is the line segment joining $(x,y)$ and $(x_0,y_0)$. Since $x\geqslant y$ and $x_0<y_0$, there is a $t\in[0,1]$ such both coordinates of $(1)$ are the same (a simple computation shows that $t=\frac{y_0-x_0}{y_0-x_0-(y-x)}$). Let $(z,z)$ be that point of the segment. Since $(x_0,y_0)$, $(z,z)$, and $(x,y)$ are colinear, with $(z,z)$ in the middle,$$\bigl\|(x,y)-(x_0,y_0)\bigr\|=\bigl\|(x,y)-(z,z)\bigr\|+\bigl\|(z,z)-(x_0,y_0)\bigr\|.\tag{2}$$But $(z,z)$ belongs to the line $y=x$ and $r$ is the distance from $(x_0,y_0)$ to that line. Therefore, $\bigl\|(z,z)-(x_0,y_0)\bigr\|\geqslant r$ and it follows from $(2)$ that $\bigl\|(x,y)-(x_0,y_0)\bigr\|\geqslant r$, which is impossible.

Therefore, if $(x_0,y_0)\in O$, $B_r\bigl((x_0,y_0)\bigr)\subset O$.