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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?

Please help me. Thanks!

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    Infinity isn't in $A$ - there is no point called infinity in $\mathbb R^2$, so there isn't such a point in $A$.2012-11-02

2 Answers 2

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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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Let's start with Problem 2:

(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?)

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    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 $S$.2012-11-02