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I really need your help.. We learned about metric spaces, open and closed sets, and I really don't understand something. Suppose I have some shape, say a 2-d square A in $\mathbb{R}^n$ without its boundary (in the intuitive sense of boundary). I want to properly prove that a point $x$ in the intuitive boundary is actually a boundary point by definition (every spherical neighbourhood of a point of the set has to contain points from both $A$ and $A^c$).

My only idea was to say that the intuitive boundary contains $x$ itself ($x$ is in $A^c$), and that it is obvious that every neighbourhood must contain some points from $A$ because we're talking about $\mathbb{R}^n$...

I don't think that's the right explanation, though. How would you prove that?

Thanks a lot!

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    $\begingroup$ You mean open and closet sets, not "groups" — right? $\endgroup$ – BrianO Nov 6 '15 at 19:42
  • $\begingroup$ Yes, of course!! Sorry, I'm not a native English Speaker and I don't study math in English and we have different terms that I tried to translate.. $\endgroup$ – PhysicsPrincess Nov 6 '15 at 20:00
  • $\begingroup$ Not a problem. I fixed it. $\endgroup$ – BrianO Nov 6 '15 at 20:18
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Lets do this in $\mathbb{R}$ for the interval $(a,b)$ (the proof is similar for the unit square or a rectangle in $\mathbb{R}^2$).

The boundary of this set is $\{a,b\}$. Why? Take a ball of radius $\epsilon$ around $a$ (which is the interval $(a-\epsilon,a+\epsilon)$). Then, $a-\epsilon/2$ is not in the interval $(a,b)$ since $a-\epsilon/2<a$. If $\epsilon < b-a$, then $a<a+\epsilon/2 < a + \epsilon < b$ so $a+\epsilon/2$ is in the interval $(a,b)$ and $(a-\epsilon,a+\epsilon)$. Similarly, if $\epsilon \geq b-a$, $(a+b)/2 \in (a,b)$ and $(a-\epsilon,a+\epsilon)$.

You can do a similar proof to show $b$ is in the boundary of this set.

If $x \in (a,b)$, $x$ is not in the boundary of $(a,b)$ since balls of radius $\min(|x-a|,|x-b|)/2$ or smaller centered at $x$ are contained entirely in the interval.

Similarly, if $x \notin [a,b]$, then $x$ is not in the boundary of $(a,b)$ since a ball of radius $\min(|x-a|,|x-b|)/2$ centered at $x$ is contained entirely outside $(a,b)$.

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For simplicity, let $A$ be the interior of the unit square in $\mathbb{R}^2$: $A = \{(x,y) \in \mathbb{R}^2\mid 0 < x < 1, 0 < y < 1\}$. Intuitively, the boundary is $B = \{(x,y) \in \mathbb{R}^2\mid (x = 0 \text{ or } x = 1) \text{ and } 0 \le y \le 1, \text{ or } (y = 0 \text{ or } y = 1) \text{ and } 0 \le x \le 1\}$.

To show that every neighborhood of each point of $B$ contains points of $A$, you have to consider each of the 4 edges that comprise B. But in fact, there are only two cases: those with some coordinate equal to $0$, and those with a coordinate equal to $1$.

For example, consider the bottom edge, excluding $(1,0)$ which can be treated when considering the right edge.

First $(0,0)$: given $\epsilon > 0$, we can assume $\epsilon < 1$. Let $r = \epsilon / 2$; then $(r,r)$ is in the open $\epsilon$-ball around $(0,0)$, and $(r,r) \in A$.

For other $(x,0) \in B$, with $0 < x$: Given $\epsilon > 0$, again we can assume that $\epsilon < 1$. Let $y = \epsilon /2$; then $d((x,y), (x,0)) = \sqrt {(\epsilon/2)^2} = \epsilon / 2 < \epsilon$, so $(x,y)$ is in the open $\epsilon$-ball around $(x,0)$, and $(x,y) \in A$.

The left edge ($x = 0$) is similar to the bottom edge ($y = 0$). The edges with one coordinate equal to $1$ are a little different in that you subtract a value to find a point that is both in $A$ and in the open $\epsilon$ ball. There are also the corner cases, literally, to consider separately.

To show that nothing else is in the boundary, two other facts need to be proved:

  • $A$ is open, and
  • $\mathbb{R}^2 \setminus (A \cup B)$ is open.

These are straightforward, and it should be fairly clear by now what their proofs are like.

In fact, this proves the proposition for any rectangle $R$ in $\mathbb{R}^2$. Suppose $R$ has its lower left corner at $\mathbf{a}$, and has width $w$ and height $h$. Then R is the image of the unit square under the homeomorphism $(x,y)\mapsto \mathbf{a} + (wx, hy)\colon \mathbb{R}^2 \to \mathbb{R}^2$. The result also follows for rectangles whose sides are not parallel to the axes, because any of those is the image under a rotation of a rectangle with sides parallel to the axes.

Proving the general statement for open rectangles in $\mathbb{R}^n$ is no different in principle — two basic cases, with subcases — but the notation gets heavy with subscripts.

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