Open tubes

olivierd

Jul 25, 2017 4 min read

In Chapter 31, “The Compact-Open Topology”, it is asserted that “clearly” the third topology on $\Mor(X, Y)$ (with $X$ and $Y$ the real line) is finer than the second. The issue is more fully discussed here.

In this post I discuss the concept of open tubes, particular subsets of the Cartesian product $X \times Y$ with $X$ a set and $Y$ a metric space.

Definition

Let $X$ be a set and $Y$ a metric space, with distance $d$ .

For any $r > 0$ and any $a \in Y$ , we note $\dot B(a, r)$ the open ball of $Y$ of radius $r$ and centred on $a$ , that is set of all $x \in Y$ such that $d(a, x) < r$ . Open balls are notoriously open subsets.

Let $\phi$ be any mapping $X \to Y$ , and $r$ a real $> 0$ .

Then the open tube $\dot T(\phi, r)$ of radius $r$ and centred on $\phi$ is the following subset of $X \times Y$ :

$\dot T(\phi, r) = \{\ (x, y) \in X \times Y\ |\ d(\phi(x), y) < r\ \}$

Equivalently:

$\dot T(\phi, r) = \bigcup_{x \in X} \{x\} \times \dot B(\phi(x), r)$

The “open” in “open tube” refers to its being based on open balls (second expression above), but does not imply that it is itself necessarily open; indeed, we have not given any topology to $X$ , so we don’t have any particular one on $X \times Y$ and cannot speak of $\dot T(\phi, r)$ being open or not.

However, if $X$ is a topological space and $\phi$ is continuous $X \to Y$ , then $\dot T(\phi, r)$ is open, as I will show.

An open tube centred on a continuous mapping is open

If $X$ is a topological space and $\phi$ is a continuous mapping $X \to Y$ , then any open tube centred on $\phi$ is an open subset of $X \times Y$ .

Proof:

I will show that in the above circumstances $\dot T(\phi, r)$ is a neighbourhood of each of its points.

Let $(x_0, y_0)$ be an eventual point of $\dot T(\phi, r)$ .

Then $d(\phi(x_0), y_0) < r$ .

Lettuce define $s = \frac 1 2 (r - d(\phi(x_0), y_0))$ . We have $s > 0$ .

Let $D = \phi^\leftarrow(\dot B(\phi(x_0), s))$ . Since $\phi$ is continuous and $\dot B(\phi(x_0), s)$ is an open subset of $Y$ , $D$ is an open subset of $X$ . Since $\phi(x_0) \in \dot B(\phi(x_0), s)$ , $x_0 \in D$ .

Let $O = D \times \dot B(y_0, s)$ , open subset of $X \times Y$ . Clearly, $(x_0, y_0) \in O$ , so $O$ is a neighbourhood of $(x_0, y_0)$ .

Let $(x, y)$ be an eventual point of $O$ .

$x \in D$ , hence $\phi(x) \in \dot B(\phi(x_0), s)$ , that is $d(\phi(x), \phi(x_0)) < s$ .

$y \in \dot B(y_0, s)$ , hence $d(y_0, y) < s$ .

$d(\phi(x), y) \le d(\phi(x), \phi(x_0)) + d(\phi(x_0), y_0) + d(y_0, y)$

Since $d(\phi(x), \phi(x_0)) < s$ and $d(y_0, y) < s$ , this leads to:

$d(\phi(x), y) < s + d(\phi(x_0), y_0) + s$

With:

$s + d(\phi(x_0), y_0) + s = 2 s + d(\phi(x_0), y_0)$
$\ = 2 (\frac 1 2 (r - d(\phi(x_0), y_0))) + d(\phi(x_0), y_0)$
$\ = r$

Thus we have: $d(\phi(x), y) < r$ .

Which implies: $(x, y) \in \dot T(\phi, r)$ .

Hence $O \subseteq \dot T(\phi, r)$ .

Thus we have found a neighbourhood $O$ of $(x_0, y_0)$ that is a subset of $\dot T(\phi, r)$ , which implies that $\dot T(\phi, r)$ itself is a neighbourhood of $(x_0, y_0)$ .

Thus $\dot T(\phi, r)$ is an open subset of $X \times Y$ . $\blacksquare$