# General way to show that a set is closed

I'm having trouble looking for a guideline on how to prove a set is open/closed:

1. Show that $A = \{(x,y) \in \mathbb{R}^{2} \mid x^{2} + y + 2x = 3\}$ is closed by showing that every limit point of A is in A.

2. a) Let $S = \{x \in \mathbb{R} \mid x \not\in \mathbb{Q}\}$, is S closed?

b) Show that $S = \{(x,y) \in \mathbb{R}^{2} \mid xy > 0\}$ is open

c) Let $A,B \subset \mathbb{R}$ with $A$ open, and defined $AB = \{xy \mid (x \in A)\wedge(y \in B)\}$, is $AB$ necessarily open?

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For example, for (1), showing that "A is closed" really means that you want to show A is a closed subset of R^2. So consider a sequence (x_n) of members of A. These members are also members of the bigger space R^2. If the sequence x_n tends to some limit point x in R as n tends to infinity, then you need to show that x is a member of the set A. That will prove that A is closed (it is a definition of "closed"). – Adam Rubinson Nov 2 '12 at 17:01
Thanks, but can you elaborate further? By looking at the graph, as n->infinity, (x,y)_n -> infinity as well. How do I prove infinity is a member of A? – user47954 Nov 2 '12 at 17:33
There isn't really a "general way" since each topology has its own set of quirks. All the topologies you are dealing with here are in Euclidean space, which usually means you need to apply properties about the real numbers and the definition of "open" and "closed" to prove whether a set is open or closed. – Thomas Andrews Nov 2 '12 at 17:34
Infinity isn't in $A$ - there is no point called infinity in $\mathbb R^2$, so there isn't such a point in $A$. – Thomas Andrews Nov 2 '12 at 17:35

To show that the set given in 1) is closed, consider a point (p,q,r) with the property that for every $\epsilon>0$, there exists a point $(a(\epsilon), b(\epsilon), c(\epsilon))$ in the set so that $d((p,q,r),(a(\epsilon), b(\epsilon), c(\epsilon)))<\epsilon$. Then, show that $(p^{2}+q+2r-3)=0$. That will complete the proof that the set is closed. You can fill in the details. A similar method works for the others.

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(a) Here $S$ is the set of irrational numbers. Does $S$ contain all of its limits points? For example, we know $0$ is a rational number and hence not in $S$. Is $0$ a limit point of $S$?
Alternatively, we could take the complement of $S$ and ask if it's open. But the complement of $S$ is just $\mathbb{Q}$. Is it true that the rationals are an open set (given the subspace topology from $\mathbb{R}$)? In other words, given a rational number, can you put an open interval around it completely contained in $\mathbb{Q}$?
(b) This is just the points $(x,y)$ in the first or third quadrant (but not on the $x$ or $y$ axis). Given such a point, can you draw an open ball around it contained entirely in its respective quadrant? (Do you see why I would pose such a question?)
To show $0$ is a limit point of $S :=$ the set of irrational numbers, you could point out that any neighborhood of $0$ will contain $1/n\sqrt{2}$ for big enough $n \in \mathbb{N}$. Of course, $1/n\sqrt{2}$ is an irrational number, so it is in $S$. – Benjamin Dickman Nov 2 '12 at 22:06