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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.
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    $\begingroup$ "Arbitrary collection" includes "finite collection"s. $\endgroup$ – Daniel Fischer Oct 14 '14 at 19:24
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    $\begingroup$ The union of an arbitrary collection of open sets being open is significantly stronger than finite unions being open, but it certainly implies it. $\endgroup$ – JHance Oct 14 '14 at 19:24
  • $\begingroup$ 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! $\endgroup$ – Nidia Oct 14 '14 at 19:28
  • $\begingroup$ 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. $\endgroup$ – copper.hat Oct 14 '14 at 19:29
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    $\begingroup$ Yes, here, 'arbitrary' if the same as 'of any kind', that is, finite or infinite. $\endgroup$ – ajotatxe Oct 14 '14 at 19:29
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By definition, in topological spaces the union of an infinite collection of open sets is open. I suspect you've been given a definition of open set in a particular space (likely a metric space such as ℝ) and asked to prove any union of open sets is open. You would have to specify the space and definition of open set to get the specific answer you desire.

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