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By a unit circle, I just mean $D = S^1$ i.e. a unit disk in $R^2$.

My game is to use some arbitrary sequence $x_n = (a_n,b_n)$ such that $b_n = \sqrt{1-(a_n)^2}$ with $0 \leq a_n \leq 1$ and show that it converges to a boundary point.

I basically feel like this is the wrong approach as I have little to no experience with analysis, and I would greatly appreciate some help.

I think the way to do this is to show that $a_n$ converges to some $0 \leq a \leq 1$ by a properties of limits, since all terms of $a_n$ are in $[0,1]$, so then $b_n$ should converge to $\sqrt{1-a^2}$ but I'm unsure whether or not the limit can be passed inside of the square root function since we haven't talked about that at all in class, and I heard somewhere mentioned that could be an issue.

Definitely though, $ 0 \leq b_n \leq 1$ is true from its definition from an arbitrary term of $a_n$, so it too should converge to a number inside $[0,1]$ so I'm not quite sure how to go about showing that it does.

Jerry Randle
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  • Why is it not enough to note that if $(a_n,b_n)$ converges to $(a,b)$ than $a_n^2+b_n^2-1$ converges to $a^2+b^2-1$? – Kapil Apr 30 '22 at 05:38
  • Because it's no guarantee under the definition I gave of $b_n$ that it converges to $b = \sqrt{1-a^2}.$ However, if you take two arbitrary sequences, there's no guarantee that they converge to the boundary either (hence the restriction on $b_n = \sqrt{(1-a_n^2)}$, although if you could somehow restrict the limits of the sequences that would work too although I'm not sure how you would do that. – Jerry Randle May 01 '22 at 21:45

1 Answers1

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Suppose $(a,b)$ is a limit point. Then there are $(a_n,b_n) \in D$ such that $(a_n,b_n) \to (a,b)$. Since $a_n^2+b_n^2 = 1$ for all $n$, and the function $(x,y) \mapsto x^2+y^2$ is continuous, we see that $a^2+b^2 = 1$ and so $(a,b) \in D$.

copper.hat
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  • Why does continuity allow us to conclude $a^2 + b^2 = 1$, given $(a_n)^2 + (b_n)^2 = 1$? Is that a particular theorem? – Jerry Randle May 01 '22 at 21:38
  • @JerryRandle Yes. Roughly it is what continuity means. – copper.hat May 01 '22 at 21:43
  • @JerryRandle In this case we have $(x_n^2+y_n^2)-(x^2+y^2) = (x_n-x)(x_n+x) + (y_n-y)(y_n+y) $, so you can see the result directly without having to think about continuity. – copper.hat May 01 '22 at 21:49
  • Oh I see, you're referring to the sequential equivalence of the definition of continuity, but you're applying it to a multivariate fcn. That makes sense, for n>>N, f($a_n, b_n$) = (a_n)^2 + (b_n)^2 converges to a^2 + b^2. Moreover, you're also using a limit location (largeness of N) theorem to say that the sum of squares sequence basically converges to 1. Neat. – Jerry Randle May 01 '22 at 21:54
  • Why is it that we can conclude $a_n^2 + b_n^2 = 1$ for all n? – Jerry Randle May 02 '22 at 20:46
  • Because they are elements of $D$. You might want to take a step back and see what the idea of the proof is. – copper.hat May 02 '22 at 20:57
  • Oh I see - I was working with the idea that elements of D are basically $x^2 + y^2 <= 1$. If D is just the ring around the unit circle (shaded in), then yes that's obvious.

    Does that set that I defined contain its limit points as well? It should, I think, given that member of that set, let's say a member of our sequence $(x_n)n$, i.e. some $x_n = (a_n, b_n)$ exactly has the restriction that $a_n^2 + b_n^2 <= 1$, and moreover $0 <= a_n^2 + b_n^2$ so by the squeeze theorem its limit L is $0 <= L <= 1$. Moreover, L is exactly $a^2 + b^2$ by the product and linearity properties of the limit.

     – Jerry Randle May 02 '22 at 21:07
  • So again, (a,b) are a member of D since $ 0 \leq a^2 + b^2 \leq 1.$ – Jerry Randle May 02 '22 at 21:14
  • You defined $D$ as $S^1$, so if $(a_n,b_n) \in D$ and $(a_,n,b_n) \to (a,b)$, then continuity (or however you want to prove it) shows that $(a,b) \in D$. – copper.hat May 02 '22 at 22:38