A set $X\subseteq\mathbb{R}$ is co-$\mathcal{W},$ if $\mathbb{R}-X$ has property $\mathcal{W}$

Question

This is a problem from Proofs and Fundamentals by Ethan D. Bloch (Problem $3.4.7$).

Suppose that $\mathcal{W}$ is some property of subsets of $\mathbb{R}.$ A subset $X \subseteq \mathbb{R}$ is $\text{co-}\mathcal{W},$ if $\mathbb{R}-X$ has property $\mathcal{W}.$ Let $\{X_i\}_{i \in I}$ be a collection of $\text{co-}\mathcal{W}$ subsets of $\mathbb{R},$ where $I$ is some indexing set.

I am asked to either prove that $\bigcup_{i \in I}X_i$ is $\text{co-}\mathcal{W},$ or give a counterexample for the following case.

1. A subset of $\mathbb{R}$ has property $\mathcal{W}$ if and only if it is finite.

My attempt: Let’s first see what kind of subsets have the property $\mathcal{W}.$ It is required that these subsets are finite. It can’t be any interval of $\mathbb{R},$ if it was, then it would be infinite.

If we have a set of $n$ real numbers, that set would be finite.

So a finite set on $\mathbb{R}$ is a set of the form $X = \{a_1, a_2, \cdots, a_n\}$ for some non-negative integer $n,$ where each $a_i$ is a real number, for all $i \in \{1,\cdots,n\}.$

By definition, $X \subseteq \mathbb{R}$ is $\text{co-} \mathcal{W},$ if $\mathbb{R}-X$ has property $\mathcal{W}.$ So, in order to have a set $X$ that is $\text{co-}\mathcal{W},$ we require $\mathbb{R}-X$ to be finite.

For example, if $X = (-\infty,0)\cup(0,\infty),$ then $\mathbb{R}-X=\{0\}.$ So $X$ is $\text{co-}\mathcal{W}.$ Note that $X$ is infinite. So the union of $\text{co-}\mathcal{W}$ subsets will be infinite. From here how can I determine, that this union is or is not $\text{co-}\mathcal{W}?$

For the next exercise, we have the case that

2. A subset of $\mathbb{R}$ is $\text{co-}\mathcal{W}$ if and only if it has at most $7$ elements. But for this one, I have no idea how to solve it.

Can someone please help me with this?

Thank you in advance!

Answer 1

1. Any union of cofinite sets is cofinite: Let $X_i \subset \mathbb{R}$ be cofinite with $Y_i := \mathbb{R} \setminus X_i$ finite. Then $\bigcap_{i \in I} Y_i$ is also finite and hence $\bigcup_{i \in I} X_i = \mathbb{R} \setminus \bigcap_{i \in I} Y_i$ is cofinite.

2. Let co-$\mathcal{W}$ be the property of having at most $7$ elements. Then a union of co-$\mathcal{W}$ sets is not necessarily co-$\mathcal{W}$. To see this take $X_1= \{1, 2, 3, 4, 5, 6, 7 \}$ and $X_2=\{11, 12, 13, 14, 15, 16, 17 \}$. Then $|X_1 \cup X_2| = 14$ and so the union is not co-$\mathcal{W}$.