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I'm trying to work through the following as part of a larger problem that I'm working on:

Invent a precise definition for the phrase "a sequence $\{x_{k}\}$ converges to infinity" in a topological space $X$. (The definition should apply, in particular, to $X=\mathbb{R}^{n}$, should not mention distance functions, and should include the phrase "for every compact subset $K\subset X$".)

I know that in a space $(X,\tau)$, a sequence $\{x_{k}\}$ is said to converge to a point $x_{0}\in X$ if, for each neighborhood $U$ of $x_{0}$, there exists an $N\in\mathbb{Z}^{+}$ such that $x_{n}\in U$ for each $n\geq N$. I also know that a "neighborhood of infinity" is a subset of $X$ which contains the complement of a closed and compact subset of $X$.

My attempt. I think the definition should read as follows.

Definition. A sequence $\{x_{k}\}$ converges to infinity in a topological space if, for each complement of a closed and compact subset $K$ of $X$, there exists a subset $U$ of $X$ containing $K^{c}$ and an there exists an index $N\in\mathbb{Z}^{+}$ such that $x_{n}\in K$ for all $n\geq N$.

I think working the definitions together is what I needed to do, but I'm not sure if this works. Does my "definition" look okay, or is there something wrong with it? Thanks in advance for any help!

Sir_Math_Cat
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3 Answers3

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Your $U$ is superfluous and could be replaces by $X\setminus K$ right away, provided $K$ is closed too:

For every compact (and closed, but this is superfluous in Hausdorff spaces) subset of $X$, there is some $N$ such that for all $n \ge N$, $x_n \notin K$.

This exactly means that $x_n \to \infty$ in the so-called Aleksandrov extension of $X$, a generalisation of the one-point compactification.

Henno Brandsma
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You can rather try "$(x_k)$ converges to infinity if for all compact $K$ of $X$, there exists a $N \geqslant 0$ such that $\forall n > N, x_n \notin K$".

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Your definition seems quite complicated. Also, I don’t understand what’s the use of $U$.

I would say: a sequence $(x_n)$ converges to $\infty$ if any compact $K$ contains only a finite number of terms of the sequence.

mathcounterexamples.net
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