Contents

# Contents

## Idea

A map $f$ between spaces (say, a continuous map between topological spaces) vanishes at infinity if $f(x)$ gets arbitrarily close to zero as $x$ gets sufficiently close to infinity.

For a map $f\colon X \to Y$, we need a notion of being close to $0$ in $Y$, so take $Y$ to be a pointed space; then getting arbitrarily close to $0$ means entering any neighbourhood of the basepoint. We also need a notion of being close to infinity in $X$, so take $X$ to be a locally compact Hausdorff space; then getting sufficiently close to infinity means entering the exterior of some compact subspace. (To interpret ‘getting’, of course, we may use nets.) It is likely, however, that further generalisations are possible.

## Definitions

Let $X$ and $Y$ be topological spaces, and let $f$ be a continuous map (or potentially any function) from $X$ to $Y$. Let $Y$ be pointed, and let $X$ be locally compact Hausdorff.

###### Definition

The map $f\colon X \to Y$ vanishes at infinity if for every neighbourhood $N$ of the basepoint in $Y$, there is compact subspace $K$ of $X$ such that $f(x)$ belongs to $N$ whenever $x$ lies in the exterior of $K$ in $X$.

In case $Y$ is a pointed metric space (such as a Banach space, with basepoint $0$; or in particular the real line, with basepoint $0$), then we may equivalently say:

• For every positive number $\epsilon$, there is compact subspace $K$ of $X$ such that ${\|f(x)\|} \lt \epsilon$ whenever $x$ lies in the exterior of $K$ in $X$.

(Here, ${\|{-}\|}$ is the norm in a Banach space, or more generally the distance from the basepoint in any pointed metric space.)

## Properties

### Relation to compactifications

One way of considering this definition is that one can adjoin to $X$ a point “at infinity”, denoted $\infty$, by declaring that the open neighborhoods of $\infty$ are sets of the form $\{\infty\} \cup Ext(K)$ for $K \subset X$ compact. This is called the one-point compactification, denoted $X^{cpt}$. Then a continuous function $f \colon X \to A$ vanishes at infinity equivalently if it extends $f$ to a map $X^{cpt} \to A$, continuous at $\infty$ (at least) that sends $\infty$ to $0$ – thus literally- “vanishing at $\infty$”.

For more see