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Fact: For any (continuous) $S^1$-action on the closed unit disk $\mathbb{D}^n$, there is a fixed point $x_0\in\mathbb{D}^n$.
I have thought of a possible argument that re-proves this, but am not sure how to complete it:

Let $U_p\subset S^1$ be the subgroup of $p^\text{th}$-roots of unity ($p$ prime). An $S^1$-action on a compact contractible space $X$ will induce a $U_p$-action on $X$. Smith Theory then implies that $X^{U_p}$ is nonempty, i.e. there is a fixed point $x_p\in X$ under $U_p$, for any given prime $p$. Now here is where I want to say: Taking $p$ sufficiently large, we find a fixed point $x_\infty$ under $S^1$. (The intuition is that $\lim_{p\to\infty}U_p\approx S^1$, and denseness will be sufficient by continuity of the action.)

1) Is it possible to fill this gap, i.e. can this 'proof' make sense? Not sure how to make sense of this limit/sequence of $U_p$'s, and whether the fixed points hop back and forth forever.

2) Is such a sequence $\lbrace x_p\rbrace_{p=\text{prime}}$ Cauchy? Or, does there exist a prime $p_0$ where $x_p=x_{p_0}$ for all primes $p>p_0$?

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# Fixed point of $S^1$-action using roots of unity

Fact: For any $S^1$-action on the closed unit disk $\mathbb{D}^n$, there is a fixed point $x_0\in\mathbb{D}^n$.
I have thought of a possible argument that re-proves this, but am not sure how to complete it:

Let $U_p\subset S^1$ be the subgroup of $p^\text{th}$-roots of unity ($p$ prime). An $S^1$-action on a compact contractible space $X$ will induce a $U_p$-action on $X$. Smith Theory then implies that $X^{U_p}$ is nonempty, i.e. there is a fixed point $x_p\in X$ under $U_p$, for any given prime $p$. Now here is where I want to say: Taking $p$ sufficiently large, we find a fixed point $x_\infty$ under $S^1$. (The intuition is that $\lim_{p\to\infty}U_p\approx S^1$)

1) Is it possible to fill this gap, i.e. can this 'proof' make sense?

2) Is such a sequence $\lbrace x_p\rbrace_{p=\text{prime}}$ Cauchy? Or, does there exist a prime $p_0$ where $x_p=x_{p_0}$ for all primes $p>p_0$?