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I apologise for this, but I have numerous potential misunderstandings.


$\Bbb R^2$ is a metric space, since it is a subspace of the Euclidean space, $\Bbb R^n$

I can look at a set $E$ with elements $x_n$ where $x_n = \frac1n,n=1,2,\dots$ and look at the plot of $n$ vs $x_n$

A limit point is a point that has another point in any of its neighbourhoods.

This means that $(n,x_n)=(1,1)$ is an isolated point, taking an $r$ neighbourhood sufficiently small, lets say $r=.1$. Isolated points include $(2,\frac12),(3,\frac13)$ so on. Is this all correct reasoning so far?

Lastly this means my only limit points are at $(\to \infty, 0)$? [and hence not in $E$]

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  • $\begingroup$ It makes no sense to say "... my only limit points are at $(\rightarrow\infty,0)$" even if you correctly point out that this is not in $E$ $\endgroup$ – Dan Rust Jan 30 '15 at 12:46
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It doesn't really make sense to say that $\mathbb R^2$ is a "subspace" of $\mathbb R^n$ (where $n>2$), because $\mathbb R^2$ is not a subset of $\mathbb R^n$. It is true that you can embed $\mathbb R^2$ in $\mathbb R^n$, by the map $(x_1, x_2)\mapsto (x_1, x_2, 0, \ldots, 0)$, but this is not really the same thing. In general, if $(X,d)$ is a metric space and $S\subset X$ is nonempty, then $(S,d')$ is a subspace, where $d'$ is the restriction of $d$ to $S$.

Assuming by your set $E$ you mean the subset of $\mathbb R^2$ consisting of the points $(n, x_n)$ where $n=1,2,\ldots$, then indeed this set has no limit points. If $(x,y)\in\mathbb R^2$, then you can take $N$ to be an integer greater than $x+1$, so for all $n\geqslant N$, $$\sqrt{(n-x)^2+\left(\frac1n - y\right)^2}\geqslant n-x\geqslant N-x > 1. $$

Your reasoning that every point of $E$ is an isolated point is correct.

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