Union of a finite number of open sets is open or not? Proper usage of this fact a proof

I received a homework assignment back and I was given full credit on the following proof:

Let $S = \{ (x,y) \in \mathbb{R}^{2} | x \geq 1$ and $y \geq 1 \}$. Is $S$ closed?

My proof is below but after reviewing a particular proposition in our book (listed below my question), I am no longer certain why I received full credit. I hope someone here can help shed light on my confusion.

Let $A = \{ (x,y) \in \mathbb{R}^{2} | x < 1 \}$ and $B = \{ (x,y) \in \mathbb{R}^{2} | y < 1 \}$. The union of these two sets is exactly the complement of $S$. That is, $\mathbb{R} \backslash S = A \cup B$.

The distance of any $x$ value for $(x,y) \in A$ to the line $x = 1$ is always $1-x$. I want an open ball (denoted $A_{r}(p)$ ) centered around a point $(x,y) = p \in A$ with radius $r>0$ to be completely contained in $A$ (to show $A$ is open). Letting $r=1-x$ would not suffice since this $r$ would guarantee points on the line $x=1$ would be included in $A_{r}(p)$. Choosing $r=\frac{1-x}{2}$ ensures that any open ball centered around some $p \in A$ is completely contained in $A$, i.e. $A_{r}(p) \subset A$. Hence $A$ is open.

A simillar argument can be made for $q \in B$ letting $r = \frac{1-y}{2} \Rightarrow B_{r}(q) \subset B$. So the set $B$ is open as well.

This next line is where I think I made a mistake.

Since $A$ and $B$ are both open, $\mathbb{R}^{2} \backslash S = A \cup B$ is open which implies $S$ is closed. QED

There is a proposition in my book that states (1) "the intersection of a finite number of open subsets of $M$ is open" and (2) "the union of an arbitrary collection of open subsets of $M$ is open.

My proof hinges on the union of a finite number of open sets being open, which seems to contradict with this proposition. So either (a) I am misunderstanding this proposition or (b) my proof is wrong.

Any insight into this would be appreciated!

• The book is Marsden, Elementary Classical Analysis 2nd ed and this proposition is on page 106.
• "Arbitrary collection" includes "finite collection"s. – Daniel Fischer Oct 14 '14 at 19:24
• The union of an arbitrary collection of open sets being open is significantly stronger than finite unions being open, but it certainly implies it. – JHance Oct 14 '14 at 19:24
• JHance, Daniel Fischer, so the intersection of an arbitrary collection of closed sets being closed would also imply that a finite intersection of closed sets is closed? Thank you both! – Nidia Oct 14 '14 at 19:28
• The meaning of arbitrary is not arbitrary. It is straightforward to show that the union of a finite number of open sets is open. If a point is in the union, it must be at least in one open set. – copper.hat Oct 14 '14 at 19:29
• Yes, here, 'arbitrary' if the same as 'of any kind', that is, finite or infinite. – ajotatxe Oct 14 '14 at 19:29