Boundary points of set $\{5-\frac{1}{n}\}$

Question

I feel rather silly for asking this but just to check....

If we have a set $P = \{ 5-\frac{1}{n} : n \in \mathbb{N}\}$

then $P = [4, 5)$

Since $\frac{1}{\infty}$ is approximately $0$ but not exactly zero.

And it would make no difference if $n$ was an element of the positive real number system $\mathbb{R}$?

Answer 1

Your intuition is right, $5-\frac{1}{n}$ gets arbitrarily close to $5$ as $n\to\infty$, so you can apply the definition of a boundary point to show that $5$ is a boundary point of $P$. As you also guessed, $4$ is a boundary point of $P$. If you are speaking of the set of boundary points of $P$, you should write $\{4,5 \}$, not $[4,5)$ because as others have pointed out, $[4,5)$ is a set containing uncountably infinite points, and it looks like you were trying to describe a set with two points. Instead of saying "$\frac{1}{\infty}$ is about $0$" I think it would be better to say $\frac{1}{n}$ is about $0$ for large enough $n$.

As for determining whether $P$ is open or closed, remember that a set is closed if it contains all its limit points. A boundary point is a type of limit point. Does $P$ contain all its limit points? You can find the closure of $P$ (denoted $\overline{P}$) with the equation $\overline{P} = P \cup P'$ where $P'$ is the set of limit points of $P$.

Answer 2

If $n$ is allowed to be a positive integer then P is the countable set {4, 4 + 1/2, 4 + 2/3, ...}. If $n$ is allowed to be a real number >= 1 then P is the uncountable set [4, 5) which includes numbers like $4 + \frac{1}{3}$ and $4 + \frac{1}{\pi}$ which are omitted in the first set.