Prove that $\{\frac 1 n \mid n \in \mathbb N\} \cup \{0\}$ is closed in $\mathbb R$
Prove that $\{\frac 1 n \mid n \in \mathbb N\} \cup \{0\}$ is closed in $\mathbb R$
I've already proved that $\{\frac 1 n \mid n \in \mathbb N\}$ is both not open and closed in $\mathbb R$ by using the points $0$ and $1$. In doing this, I've used the Euclidean distance as the metric. Is it right to do so, or should this be proved using an arbitrary metric ? (please comment on this - the exercise mentions nothing about the metric on $\mathbb R$)
I know I can prove $A=\{\frac 1 n \mid n \in \mathbb N\} \cup \{0\}$ is closed by proving that for every convergent sequence $\{a_n\}$ in $A$ we have $\lim_{n\rightarrow \infty} a_n \in A$.
On the other hand how can $A$ be closed ? Let $b$ be an irrational number in $[0,1]$. How can we find $r > 0$ such that $B_r(b) \subseteq A^C$ ?
Solution 1:
More elementary:
Let $x \in \mathbf{R}\setminus A$. If $x < 0$ or $x > 1$, finding a ball centered at $x$ and contained in $\mathbf{R}\setminus A$ is trivial. If $0 < x < 1$, then $1/(n+1) < x < 1/n$ for some $n$, this gives you an open interval of $\mathbf{R}\setminus A$ of non-zero length which contains $x$, in which you can fit a ball centered at $x$.
Solution 2:
Hints assuming the usual topology on $\;\Bbb R\;$:
1) A subset of the reals is closed iff it contains all its limit points;
2) If a real sequence converges then any subsequence also converges and to the same limit.
Solution 3:
Twisted proof: prove directly that the set is compact. Given a open cover, the open set containing 0 will contain almost all the elements of the set.
Solution 4:
Notations:
$A^C$ means the complement of $A$, i.e. $A \cup A^C = U$. In general, $U = \mathbb{R}$ unless there is a specific explanation.
$(s_n)$ means a sequence. For example, $(\frac{1}{n})$ means the sequence $(1, 1/2,1/3, 1/4, 1/5, 1/6,...)$, i.e. $s_n = \frac{1}{n}$ where $n \in \mathbb{N}$.
Proof:
Let $A = \{\frac{1}{n}: n \in \mathbb{N}\} \cup \{0\}$. For proving $A$ is a closed set, it equals to prove $A^C$ is an open set. Observing that the sequence $s_n = \frac{1}{n}$ ($n \in \mathbb{N}$) forming a strictly decreasing sequence, hence, we could construct a bunches of sets such that for set $S_n$: $\sup{S_n}=s_n=\frac{1}{n}$ and $\inf S_n = s_{n+1} = \frac{1}{n+1}$, $\sup S_n \notin S_n$ and $\inf S_n \notin S_n$, then no element from the sequence $s_n$ would be in the sets $S_n$.
Therefore, $A^C = \cup_{n=1}^{+\infty} (s_{n+1},s_n) \cup (-\infty,0) \cup (1,\infty)$.
Because the union of any collection of open sets is an open set, clearly, each $S_n$ is an open set and $(-\infty,0)$ and $(1,\infty)$ are open sets. Therefore $A^C$ is open [1] (Contrastively, the intersection of finitely many open sets is an open set). Since $A = \mathbb{R} - A^C$ and $A^C$ is an open set, we proved that $A$ is a closed set by the definition of closed set.
Reference:
[1] Elementary Analysis: The Theory of Calculus, 2nd Edition. Kenneth A. Ross. Page 87, discussion 13.7, Springer, 2013.