How do I find the interior, closure and boundary of an arbitrary set?

Question

I'd like to find the interior, boundary, closure, etc. of an arbitrary set. I mean, if I'm working on a subset of $\mathbb{R}, \mathbb{R}^2,$ or $\mathbb{R}^3$ it could be quite easy using Geogebra, but in other cases or sets that are complex to graph, it turns out to be very difficult.

So I'm asking about an algebraic method (most specifically, an example to understand) to find those sets.

Like $A=\{(x,y)\in \mathbb{R}^2:0<(x-2)^2+y^2<1\}\cup\left\{\left(\dfrac{1}{n}+2,0\right):n\in\mathbb{N}\right\}$

Answer 1

In general, there is no procedure for finding the interior, boundary, and closure of a set, even in $\mathbb{R}^2$. As an example of some weirdness you could encounter, consider the set $$ S= \Big\{ \Big(x,\sin\Big(\frac{1}{x}\Big) \Big) \colon x > 0 \Big\} \subseteq \mathbb{R}^2. $$ Then for each value $z \in [-1,1]$ we can find infinitely many distinct terms $0 < x_n < 1$ such that $\sin(\frac{1}{x_n}) = z$. It follows that the sequence $(x_n, \sin(\frac{1}{x_n}))$ converges to $(0,z) \in \mathbb{R}^2$. Since $(0,z)$ is a limit point of $S$, by definition $(0,z)$ is in the boundary of $S$ for all $z \in [-1,1]$. You wouldn't necessarily guess that by looking at the formula for $S$.

Of course it's possible to construct even more pathological examples where the boundary is even harder to determine. See for example the Mandelbrot set.

Edit: Just to throw in another weird example the Sierpiński Triangle is a closed set with empty interior. It is therefore equal to its own boundary.

Answer 2

You have to develop intuition about what it means to be a limit point, on the boundary, etc. Sometimes looking at a graph is helpful. Here's a sketch of the set. Note that all but one of the sequence of points converging to $(2, 0)$ are already in the open punctured disk, so the set could be described more simply as: $$ A = \{ (x,y) \in \mathbb{R}^2: 0 <(x-2)^2 + y^2 < 1 \} \cup \{ (3, 0) \}. $$ Here's a sketch, where I've indicated the sequence of points so you can see them:

[![A set of points in the plane][1]][1]

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To answer your questions, the algebraic inequalities used in defining the sets allow you to rigorously confirm the following conclusions:

1. The interior of the set consists of all points who have an open neighborhood (e.g. a disk of arbitrarily small size) around them in $A$. This is just the punctured disk: $$ A^\circ = \{ (x,y) \in \mathbb{R}^2: 0 <(x-2)^2 + y^2 < 1 \} $$
2. The closure includes the set and all of its limit points. The limit points are not necessarily in the set, but any neighborhood around a limit point intersects the set. You get the center of the disk and the circle of points at the boundary of the disk: $$ \overline{A\,} = \{ (x,y) \in \mathbb{R}^2: (x-2)^2 + y^2 \leq 1 \} $$
3. The boundary consists of all points who have a neighborhood that intersects both the set and its complement, or equivalently the closure with the interior removed. This consists of the center of the disk and the circle: $$ \partial A = \{ (2,0) \} \cup \{ (x,y) \in \mathbb{R}^2: (x-2)^2 + y^2 = 1 \} $$ [1]: https://i.stack.imgur.com/vLWXH.png