Boundary of a set in metric space

Question

I'm trying to find the boundary of a set in order to check whether or not the set is open or closed on the metric I'm given. While I understand the latter, I'm not sure how to properly understand the former. The set is given by $$M=\left\{\vec v = \begin{pmatrix}v_1 \\v_2 \\\end{pmatrix} \in \mathbb{R}^2 \ \middle|\ -1 \leq v_1 < 1 \right\}. $$ Am I right in assuming that my set is only bounded to the left and right by $[-1,1)$ and approaches infinity when going "upward" and "downward"? How would I describe the boundaries, then? Thanks for any help or tips.

Answer 1

As you say, the set is a vertical band in $\mathbb{R}^2$. The picture looks like this:

The boundary of a set consists of all of the points that are "close to" the set and also "close to" the complement of the set, where "close to" is a term that I will make rigorous below. From the picture, it appears that the boundary consists of two vertical lines: $\{x=-1\}$ and $\{x=1\}$.

Now, to make it rigorous:

Definition: Let $E \subseteq \mathbb{R}^2$ (or, more generally, let $E$ be a subset of a metric space). A point $x$ is a boundary point of $E$ if for any $\varepsilon > 0$, we have $$ B(x,\varepsilon) \cap E \ne \varnothing \qquad\text{and}\qquad B(x,\varepsilon) \cap E^\complement \ne \varnothing.$$

In other words, a point is a boundary point if every ball around that point intersects both the set itself, and the complement (that is, all of the points in the boundary are close to both the set and its complement). In the case of the set $M$ which you define, any ball around a point of the form $(\pm 1, y)$ contains some point in the set, and some point outside of the set, e.g., assuming that $\varepsilon < 2$, $$ \left(1-\frac{\varepsilon}{2}, 0\right) \in B(1,0) \cap M, \qquad\text{and}\qquad \left(1+\frac{\varepsilon}{2}, 0\right) \in B(1,0) \cap M^\complement. $$

On the other hand, if you take any other point $x$ in the plane, then we can find a $\varepsilon > 0$ so small that the ball of radius $\varepsilon$ centered at that point is either completely in $M$ (if $x\in M$), or in the complement of $M$ (if $x\in M^\complement$).

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This set is neither open nor closed. To make sense of this, we require a definition of what it means for a set to be open or closed. There are many equivalent definitions; I'm going to go with the following:

Definition: A set $U$ is open if for any $x\in U$ there exists some $r > 0$ such that $B(x,r) \subseteq U$. A set $K$ is closed if it is the complement of an open set.

In the case of $M$, above, it is not open. To see this, note that $(-1,0)$ is in $M$, but no ball around this point is completely contained in $M$ (this was noted above). On the other hand, $M^\complement$ is not open either, since no ball around $(1,0)$ is completely contained in $M^\complement$.

This, by the way, suggests another notion of "closed":

Definition: A set is closed if it contains all of its boundary points. A set is open if it is the complement of a closed set.

Hence we can determine the openess or closedness of a set by examining its boundary, which is likely what this exercise was intended to get at in the first place. ;)