Open tubes

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

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