Intereting Posts

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Proving pointwise convergence of series of functions
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Complex Conjugate of Complex function
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The exercise is to find the accumulation points of the set $S=\{(\frac {1} {n}, \frac {1} {m}) \space m, n \in \mathbb N\}$

I’m trying to prove that if $A$={accumulation points of the set $S$}, then $A=\{(0,0), (\frac {1} {n},0),(0,\frac {1} {m}) \space n,m \in \mathbb N\}$. I could prove that the set of the right side of the equality is included in $A$. I don’t know how to prove the other inclusion, which means that if $x$ is an accumulation point of $S$, then $x$ has to be $(0,0)$, or of the form $(\frac {1} {n},0)$ or $(0,\frac {1} {m})$.

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Find the distance between points $\left(\frac{1}{n}, \frac{1}{m}\right)$ and $\left(\frac{1}{n’}, \frac{1}{m’}\right)$. Use an argument based on Cauchy sequences to conclude that any Cauchy sequence of $S$ (that isn’t on the diagonal) will eventually only vary in one of its components. Find the limit.

It’s been a loooong time since I did this problem (10 years?) but that should at least be a big hint.

Here’s another solution: let $$B=\{(\frac1n,0)\mid n\in\mathbb N\}\cup\{(0,\frac1n)\mid n\in\mathbb N\}\cup\{(0,0)\}.$$ Now, note that horizontal and vertical strips of the form $(a,b)\times\mathbb R$ and $\mathbb R\times(c,d)$ are open sets. The complement of $B\cup S$ is a union of such strips. Therefore, $B\cup S$ is closed, so it must contain every accumulation point of $S$. To complete the proof, it remains to verify that each point in $S$ is isolated (since we already know that every point in $B$ is an accumulation point).

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