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I have having trouble with some simple problems involving the closure of sets. From my understanding the closure of a set is a set that contains all possible limits

a) $\{ (x,y) \in \mathbb{R^2} : xy< 1 \}$

b) $\{ (x,\sin(\frac{1}{x}): x > 0 \}$

c) $\{(x,y) \in \mathbb{Q^2}: x^2 + y^2 < 1 \}$

Here are is my attempt

a) $\{ (x,y) \in \mathbb{R^2} : xy \leq 1 \}$

b) Just itself? Not sure if this is even remotely meaningful

c) I think this is just the unit disk, not just why being members of only rationals makes any difference

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up vote 1 down vote accepted

I like the definition of closure as the smallest closed set containing the set.

a) You are correct.

b) You should plot this, particularly near $x=0$ where $\sin \frac{1}{x}$ oscillates wildly. The points $(\frac{1}{(2n+1) \frac{\pi}{2}}, (-1)^n) $ are in the set and $\sin$ is continuous, hence you have the closure $\{ (x,\sin(\frac{1}{x}): x > 0 \} \cup \{0\} \times [-1,1]$

c) Any closed set which contains the open disk must also contain the edge of the disk. Hence any closed set containing the disk contains the closed set $\{(x,y) \in \mathbb{R}^2 | x^2+y^2 \leq 1\}$, hence this is the closure.

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For (b) why is {0} also in the closure? Why do I have to include -1 and 1 if I just let $x \to 2/\pi$? Isn't that already in the set itself? – Hawk Oct 22 '12 at 2:27
Well, $(0,\pm 1)$ must be there since they are limit points of the sequence of points I described above. Furthermore, since $x \mapsto \sin \frac{1}{x}$ is continuous, all points in between must also be limit points. And this includes $(0,0)$. Look at Brian's curve, it will be clear... – copper.hat Oct 22 '12 at 3:13
So then wouldn't {0} x [-1,1] (the infinite "horizontal" rectangle) be sufficient? – Hawk Oct 22 '12 at 15:21
The $\{0\} \times [-1,1]$ is a line segment. You still need the rest of the curve (for $x>0$). – copper.hat Oct 22 '12 at 15:27
Oh {0} x [-1,1] you were referring to the line x = 0 with y = t with $t\in [-1,1]$? – Hawk Oct 22 '12 at 15:53

You've got (a) and (c) (provided we construe "unit disk" as "closed unit disk").

For (b), notice that every point of the form $(0,y)$ for $|y|\le 1$ is in the closure. The function $x\mapsto \sin(1/x)$ oscillates infinitely many times between $-1$ and $1$ for $x$ between $0$ and any positive number. So, for example, you get an infinite sequence $x_1,x_2,x_3,\ldots$ approaching $0$ such that $\sin x_1=\sin x_2=\sin x_3=\cdots = 3/4$. So $(0,3/4)$ is in the closure.

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The closure of a set $A$ is $A$ together with any limit points of $A$ that aren’t already in $A$. If $A$ is the set in (a), the points on the hyperbola $xy=1$ are indeed the only limit points that aren’t already in $A$, so your answer is correct.

In (c) the closure of $A$ is indeed the closed unit disk: not only are the points on the unit circle limit points of $A$, but because the rationals are dense in the reals, every point inside the unit circle is a limit point of $A$ as well.

The set in (b) is just the graph of the function $f(x)=\sin\frac1x$ over the domain $x>0$. There’s a picture of it here; see how it squeezes towards the $y$-axis, or at any rate the part of the $y$-axis between $y=-1$ and $y=1$? That should suggest a bunch of limit points of $A$ that aren’t already in $A$.

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