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I was able to determine (c) is compact since $(e^{-x}\cos(x),e^{-x}\sin(x))$ spirals back to the x-axis and the horizontal segment bounds the area. But does the curve keep spiralling making it unbounded outside the horizontal segment?

For (d), I tried to plot the parametric surface in Mathematica, but all I got is a blank plot. Is there an algebraic way to do this? My best attempt was (let $x = u$ and $\theta = v$)

$x^2 + y^2 = e^{-2u}(\cos^2v + \sin^2v) = e^{-2u} \to 0$

EDIT: For (c), wouldn't it be not compact since it isn't the intersection, but the union?

EDIT2: Let me clarify. If it was the intersection of the two sets, I get only the point where they meet and that is only one point. If it were the union, I get the set of all points on the curve and the first set is clearly not closed (as demonstrated in (d) ). So I don't think the set is compact. Am I wrong?

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(d) is not closed. You can get as close as you want to the origin by making $x$ large enough. Yet, the origin is not a point of the set.

By the way, here is how to plot (d) via Mathematica:

ParametricPlot[{Exp[-x] Cos[t], Exp[-x] Sin[t]}, {x, 0, 5}, {t, 0, 2 Pi}] 

To see how $(e^{-x}\cos\theta, e^{-x}\sin\theta)$ changes as $x$ varies, try:

Manipulate[ParametricPlot[{Exp[-x] Cos[t], Exp[-x] Sin[t]}, {t, 0, 2 Pi}, PlotRange -> {{-2, 2}, {-2, 2}}], {x, 0, 5}] 
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    Why are you making x large? I can get to the origin if it x is small2012-10-30
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    @jak I'm making $x$ large to make the radius of the circle $(e^{-x}\cos\theta, e^{-x}\sin\theta)$ small, which makes it approach the origin. $e^{-x}$ decreases as $x$ increases.2012-10-30
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    I plotted the same thing, but I wasn't able to read what the heck it is, so then I noticed the ordered paid has two variables. That led me to plot it in 3D2012-10-30
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    @jak I added another code snippet. Give it a try. I think it'll show you how the value of $x$ affects (d).2012-10-30
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    I know what you mean. But why are we looking at just $x$? Because $\theta$ can make the point go to the origin anyways2012-10-30
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    @jak No value of $\theta$ can make both $\cos\theta$ and $\sin\theta$ equal to $0$. In fact, the distance from the origin doesn't change if you change $\theta$ alone.2012-10-30
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    okay I think I see your point. It applies to (d), I don't see how it applies to (c) though. The horizontal segment covers the origin no?2012-10-30
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    @jak You're right. I missed the horizontal segment part. (c) is closed and bounded, hence compact.2012-10-30
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    I think we made a mistake. For (c) shouldn't it be the intersection, not the union?2012-11-01
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    @jak Our discussion is correct when considering the union. The intersection, on the other hand, isn't compact because it isn't closed (doesn't contain the origin, which is still a limit point).2012-11-01