Since last week I've been learning a bit about Topology in Calculus and know the basic definitions of open, closed, norm, etc. Now I try to solve this question but I don't know how to. Its really really confusing and difficult to imagine...
Let $(X,d) = (\mathbb{R^2}, d_\infty)$ and $M := [0, \infty) \times (0,\infty) \subset X.$ Prove:
- $\quad A_1 := \left \{{(x, y) \in \ X : 0 \leq x < y}\right \}$ is open in $M$, but not in $X$
- $\quad A_2 := \left \{{(x, y) \in \ X : 0 < y \leq x}\right \}$ is closed in $M$, but not in $X$
Can someone help me?
Thanks
Landau
I Believe that M be with the induced topology. Remark that $U=\{(x,y)\in\mathbb{R}^2;\, x<y\}$ is open in $X$ and too $F=\{(x,y)\in\mathbb{R}^2;\, y\leq x\}$ is closed in X, with $U^c=F$. In fact, $f:X\to \mathbb{R}$, $f(x,y)=x-y$ is continuous, then look that $U=f^{-1}((-\infty,0))$ and $F=f^{-1}([0,+\infty))$.
Now, remark that $A_1=U\cap M$ and $A_2=F\cap M$. Thus $A_1$ is open in $M$ and $A_2$ is closed in $M$.
If $A_1$ is open in $X$, then exists a ball (or square) $B$ centered at $(0,1)\in A_1$ such that $B\subset A_1$. Thus, exists $(x,y)\in A_1$ with $x<0$, which is a contradiction with the definition of $A_1$. The case of $A_2$ note that $(1,0)\in cl(A_2)\setminus A_2$, then $A_2$ is not closed in $X$.