# Prove that a sequence $\{x_k\}_{k=1}^\infty\subset \mathbb{R}^n$ converges to $x$ if and only if the map $f(j) = x_j$ is continuous.

Need to know how to prove that a sequence $\{x_k\}_{k=1}^\infty\subset \mathbb{R}^n$ converges to $x$ if and only if the map $f:\{1,2,3...\} \to\mathbb{R}^n$, $f(j) = x_j$, is continuous.

It's been puzzling me for the last few hours now. I know that a sequence converges to a point just when every open set containing x contains all but finitely many of the points in the sequence.

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Please check that I didn't change the meaning. – Dennis Gulko Nov 13 '12 at 10:40
The set $\{1,2,3...\}$ is discrete, so any map $f:\{1,2,3...\} \to\mathbb{R}^n$ is continuous. – Dennis Gulko Nov 13 '12 at 10:42
This is an odd question and I don't think it's true: if you take the naturals $\,\{1,2,3,...\}\,$ with the inherited topology from the reals, then this is a discrete space, but any function from a discrete space to any other space is always continuous... – DonAntonio Nov 13 '12 at 10:43
No, that's not a proof, because for instance the function $f:\Bbb{N}\to\Bbb{R}$ given by $f(n)=n$ is continuous but corresponds to the sequence $\{n\}$, which doesn't converge. The claim is false, although it might work with the cofinite topology on $\Bbb{N}$. – Kevin Carlson Nov 13 '12 at 10:58
It won't work with to cofinite topology either, I think. The statement "$x_k \to x$" depends on $x$, but "$f \colon j \mapsto x_j$ is continuous" doesn't. – martini Nov 13 '12 at 11:42

In the comments it is pointed out that this is definitely false when $\mathbb{N}$ is equipped with the discrete topology. Following Kevin's suggestion, one might also consider the cofinite topology on $\mathbb{N}$, that is, the topology whose open sets are $\varnothing$ and complements of finite sets. This still does not work. In fact:

Let $\mathbb{N}$ have the cofinite topology, and let $\{x_k\}_{k=1}^\infty$ be a sequence in $\mathbb{R}^n$. Then $k\mapsto x_k$ is a continuous map $\mathbb{N}\to \mathbb{R}^n$ if and only if $\{x_k\}_{k=1}^\infty$ is a constant sequence.

Proof: The "only if" direction immediate, since constant functions are always continuous. So now assume $\varphi\colon \mathbb{N}\to \mathbb{R}^n$ given by $k\mapsto x_k$ is continuous. Since $\mathbb{N}$ is compact in the cofinite topology, the set $\{x_k : k\in \mathbb{N}\}$ is compact in $\mathbb{R}^n$. In particular, the sequence $x_k$ has a limit point $x$. For each $\epsilon>0$, let $B_\epsilon$ be the closed ball around $x$ of radius $\epsilon$. Then $\varphi^{-1}(B_\epsilon)$ is a closed, infinite subset of $\mathbb{N}$. The only closed, infinite subset of $\mathbb{N}$ in the cofinite topology is $\mathbb{N}$. Thus $\varphi^{-1}(B_\epsilon) = \mathbb{N}$ for each $\epsilon>0$, which says precisely that $x_k = x$ for all $k$.

This proves the cofinite topology doesn't work. In fact, nothing works:

There is no topology on $\mathbb{N}$ such that a sequence $\{x_k\}_{k=1}^\infty$ in $\mathbb{R}^n$ converges if and only if $\varphi\colon \mathbb{N}\to \mathbb{R}^n$ given by $k\mapsto x_k$ is continuous.

Proof: Assume such a topology existed. Let $A\subseteq\mathbb{N}$ be any subset. Define a sequence $x_k\in \mathbb{R}^n$ by $$x_k = \frac{\eta_k}{k}e_1,$$ where here $e_1$ the first standard basis element of $\mathbb{R}^n$, and $\eta_k$ is $1$ is $k\notin A$ and $-1$ if $k\in A$. This sequence converges to $0\in \mathbb{R}^n$, and hence by assumption $\varphi(k) := x_k$ is continuous $\mathbb{N}\to \mathbb{R}^n$. The set $U = \{y\in \mathbb{R}^n : y_1<0\}$ is open, so $\varphi^{-1}(U)$ must be open in $\mathbb{N}$. Note that $\varphi^{-1}(U)$ is exactly $A$, by construction. We have thus shown that any subset $A\subseteq\mathbb{N}$ is open, or in other words, the topology on $\mathbb{N}$ must be the discrete topology. However, we have seen that the statement fails for the discrete topology, and therefore there is no topology on $\mathbb{N}$ that works.

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