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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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    @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