# Closure, interior..

• Apr 29th 2010, 03:25 PM
Julius
Closure, interior..
1. Let $\displaystyle A=\{x\in R^3:x_2\geq 2, x_1^2+x_3^2=1\}$
Find $\displaystyle A^\circ,\overline{A}$ and explain.
id say that interior is empty set and closure is A, but if so, how to show that?
2. Let $\displaystyle A=\{\|x\|_{max}\leq 2,x\in\mathbb{R}^2\}$ and $\displaystyle B=\{x^2-y^2>1,(x,y)\in\mathbb{R}^2\}$. Let $\displaystyle C=A\cap B$. i dont know how exactly its called but: is C partly opened with respect to A, or partly closed?

definition: Y is partly opened with respect to X if and only if there exists opened set $\displaystyle G\in\mathbb{R}^d$ such that $\displaystyle Y=G\cap X$
• Apr 29th 2010, 04:51 PM
Drexel28
Quote:

Originally Posted by Julius
1. Let $\displaystyle A=\{x\in R^3:x_2\geq 2, x_1^2+x_3^2=1\}$
Find $\displaystyle A^\circ,\overline{A}$ and explain.
id say that interior is empty set and closure is A, but if so, how to show that?

Hint:
Spoiler:

Call $\displaystyle \pi_2$ the canonical projection onto the second coordinate space. and $\displaystyle \gamma:\mathbb{R}^3\to\mathbb{R}^2:(x_1,x_2,x_3)\t o (x_1,x_3)$. Clearly the first map is continuous and the second is because each of it's coordinate functions is continuous. So then $\displaystyle A=\pi_1^{-1}\left([2,\infty)\right)\cap\gamma^{-1}\left(\mathbb{S}^1\right)$ (where $\displaystyle \mathbb{S}^1$ is the unit circle). So what?

Quote:

2. Let $\displaystyle A=\{\|x\|_{max}\leq 2,x\in\mathbb{R^d}\}$ and $\displaystyle B=\{x^2-y^2>1,(x,y)\in\mathbb{R^d}\}$. Let $\displaystyle C=A\cap B$. i dont know how exactly its called but: is C partly opened with respect to A, or partly closed?

definition: Y is partly opened with respect to X if and only if there exists opened set $\displaystyle G\in\mathbb{R}^d$ such that $\displaystyle Y=G\cap X$
What do you think?
• Apr 30th 2010, 12:40 AM
Julius
Quote:

So what?
$\displaystyle A^\circ=X^\circ\cap Y^\circ$, $\displaystyle \gamma^{-1}(\mathbb{S}^{1})^\circ=\emptyset$ so $\displaystyle A^\circ=\emptyset$?

Quote:

What do you think?
i dont know.. B isnt closed, so C wont be closed, $\displaystyle X=\{x^2-y^2\leq 1,(x,y)\in\mathbb{R}^2\}$, $\displaystyle R^2/ X$ is open so C is open?
• Apr 30th 2010, 07:01 AM
Drexel28
Quote:

Originally Posted by Julius
$\displaystyle A^\circ=X^\circ\cap Y^\circ$, $\displaystyle \gamma^{-1}(\mathbb{S}^{1})^\circ=\emptyset$ so $\displaystyle A^\circ=\emptyset$?

Well, first-off it tells you that $\displaystyle A$ is the intersection of the preimages of two closed sets under a continuous map, in other words it's the intersection of two closed sets, in other words it's closed, in other words $\displaystyle \overline{A}=A$

Why do you think $\displaystyle \left(\gamma^{-1}\left(\mathbb{S}^1\right)\right)^{\circ}=\varnot hing$?

Quote:

i dont know.. B isnt closed, so C wont be closed, $\displaystyle X=\{x^2-y^2\leq 1,(x,y)\in\mathbb{R}^2\}$, $\displaystyle R^2/ X$ is open so C is open?
I don't understand I guess what's the problem. What you're calling "party open" just means that it's open in the subspace topology (or open as a metric subspace...whatever your prefer), in other words it's the intersection of $\displaystyle A$ with an open set. But, once again $\displaystyle \varphi:\mathbb{R}^2\to\mathbb{R}:(x,y)\mapsto x-y$ is continuous and $\displaystyle B=\varphi^{-1}\left((1,\infty)\right)$...soooo?