Answer to Question #267262 in Real Analysis for K123

Question

Tell whether the following sets A are (i) open or closed, (ii) connected,

(iii) compact. Find (iv) the limit points of A, (v) int(A), (vi) the boundary of A.

A= rational numbers

A = integers

Answer 1

First let us start with the case $A= \mathbb{Q}\subset \mathbb{R}$

1. $A$ is neither open (as any neighbourhood of any point contains an irrational number) nor closed (as a limit of rational numbers need not to be rational).
2. $A$ is not connected, we can consider two following sets : $A^1 = (-\infty;\sqrt{2}), A^2 = (\sqrt2 ; +\infty)$. Both of them are open and $A\subseteq A^1\cup A^2$.
3. $A$ is not compact, as it is not closed in $\mathbb{R}$ (any compact subset of $\mathbb R$ is, in particular, a closed subset).
4. Limit points of $A$ is the entire real line $\mathbb{R}$, as any real number is a limit of a sequence of rational numbers.
5. $\text{int} A = \empty$, as any non-empty open set contains an irrational number, and so no non-empty open set is contained in $A$.
6. $\partial A = \mathbb{R}$, as the boundary is exactly the closure of $A$ minus the interior of $A$, i.e. $\mathbb{R} \setminus \emptyset = \mathbb R$.

Now let us consider the case $A = \mathbb Z \subseteq \mathbb R$

1. $A$ is closed, as any converging sequence of integers should be stable, so it converges to an integer. $A$ is not open, as any neighbourhood of an integer contains non-integer numbers.
2. $A$ is not connected, we can consider the sets $A^1=(-\infty; \frac{1}{2}), A^2=(\frac{1}{2}, +\infty)$. Both of them are open and $A\subseteq A^1 \cup A^2$.
3. $A$ is not compact, as the sequence $x_i = i$ does not admit a converging subsequence. Alternatively, the open cover $(i-\frac{1}{2}, i+\frac{1}{2})$ for $i\in \mathbb Z$ does not admit a finite subcover.
4. The limit points of $A$ are empty, as no integer $n \in \mathbb Z$ can be approximated by a sequence of other integers.
5. $\text{int} A = \empty$ as any non-empty open set contains a non-integer number.
6. $\partial A = A = \mathbb{Z}$, as $A$ is a closed set with an empty interior.