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If we denote the general point of $\mathbb{R}^2$ by $(x,y)$, determine $\operatorname{Int}A$, $\operatorname{Ext}A$, and $\operatorname{Bd}A$ for the subset $A$ of $\mathbb{R}^2$ specified by each of the following conditions

(a) $x = 0$

(b) $0 \leq x < 1$

For (a), could $\operatorname{Bd}A= \varnothing$? Because I can't decide whether $x = 0$ is open or closed. I have that $\operatorname{Int}A= A$

I am also having trouble writing my $\operatorname{Ext}A$ in a compact way. The following is what I have

$$\{ (x,y) \in \mathbb{R}^2:-\infty < x < 0\} \cup \{ (x,y) \in \mathbb{R}^2:0 < x < \infty\} $$

How do I write this more compactly? Perhaps using cross product?

(b) For $\operatorname{Ext}A$, same problem. I had to write it as a union of two disjoint sets. For $\operatorname{Int}A$, is it okay to just leave as $\operatorname{Int}A = \{(x,y) \in \mathbb{R}^2: 0 < x < 1\}$? Can I assume that the reader knows $-\infty < y < \infty$?

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For a , I think we should consider $A=\{(x,y)\mid x,y\in\mathbb R, x=0\}$ instead. Shouldn't we?? – Babak S. Jan 15 '13 at 5:55
What do you mean? – Hawk Jan 15 '13 at 5:57
I mean $x=0$ should epitomize a subset in $\mathbb R^2$. What $x=0$ mean in the Cartesian plane? – Babak S. Jan 15 '13 at 6:00
Are you asking me what $x = 0$ it is in $\mathbb{R}^2$? It's the y-axis – Hawk Jan 15 '13 at 6:02
I wrote it as $A = \{ (0, y) \in \mathbb{R}^2\}$ – Hawk Jan 15 '13 at 6:05
up vote 2 down vote accepted

For (a), the boundary is not empty. Every point in $A$ where $x=0$ is adjacent to points in $A$ and points not in $A$, informally speaking. So Bd A = $\{(x,y)~|~x=0\}$. The interior of the set is what is empty. The exterior you can write more simply $\{(x,y)~|~x\neq0\}$

Part (b) you can do the same thing in describing the exterior. Your description of the interior looks good to me and you can indeed assume $-\infty<y<\infty$ because that is inherent in the definition of $\mathbb{R}^2$.

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For (a), that's what I am confused about. Why isn't the interior the same as $A$? I can certinaly draw a neighborhood in A such that it is still contained in the set no? And why do you say the boundary is 0 and not 0? – Hawk Jan 15 '13 at 6:03
@sizz: No, you can’t: any nbhd of a point $\langle 0,y\rangle$ must contain points with non-zero $x$-coordinate. – Brian M. Scott Jan 15 '13 at 6:05
@BrianM.Scott, how? I am fixing $x = 0$, how can I escape from $A$? I don't see how the nbhd can contain other x-points – Hawk Jan 15 '13 at 6:06
@sizz How does your text/notes define interior and boundary of a set? One way to define it (not sure if it is the one you should use in your class) is that you can draw a ball around an interior point that is completely contained in $A$, whereas any ball drawn around a boundary point, no matter how small, contains points in $A$ and points not in $A$. – RussH Jan 15 '13 at 6:08
@sizz: Yes: you’re working in $\Bbb R^2$ with its usual topology, not just in the $y$-axis. – Brian M. Scott Jan 15 '13 at 6:11

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