Continuity in Metric Spaces
Following the case of real analysis, let’s define continuous functions via the usual ε-δ definition.
Definition. Let (X, d) and (Y, d’) be two metric spaces. A function f : X → Y is said to be continuous at if:
- for each ε>0, there exists δ>0 such that whenever satisfies d(x, a) < δ, we have d’(f(x), f(a)) < ε.
Equivalently, we can replace the condition “whenever … ” by the set-theoretic statement which is often more convenient.
If f is continuous at every a, we just say that f is continuous.
Theorem. The function f : X → Y is continuous at a iff for any open subset containing f(a), there’s an open subset containing a.
In particular, f is continuous if and only if for any open subset , the pull-back is open.
Suppose f is continuous at a. If V is an open subset of Y containing f(a), then for some ε>0. By continuity at a, there exists δ>0 such that . Hence is an open subset containing a.
For the converse, suppose the stated condition holds. Given ε>0, is an open subset of Y containing f(a), so there is an open containing a. Since U is open, there’s an open ball . It thus follows that .
For the second statement, suppose f is continuous and is open. By what we just proved, each element a of is contained in an open subset which is contained in U. Thus U is a union of open subsets and is hence open.
Conversely, suppose whenever is open, so is Let a be any point in X. From what we just proved, f is continuous at a. ♦
The theorem tells us that continuity is fundamentally a statement on the underlying topologies. In other words, if f : (X, d) → (Y, d’) is a continuous map of metric spaces, then we can replace d or d’ by any topologically equivalent metric and it wouldn’t make any difference.
Next, we prove that the metric itself is continuous (considering this map is so fundamental, it’d be pretty weird if it weren’t).
Proposition. In a metric space (X, d), the map is continuous.
First, recall that there’s no fixed way of defining a metric on X × X, so we pick one of the many topologically equivalent ones, say
Let . To show continuity at that point, suppose ε>0. Letting δ = ε/2, we see that whenever , we have: d(x, x’) < δ and d(y, y’) < δ and thus the triangular inequality gives:
- d(x, y) – d(x’, y’) ≤ d(y’, y) + d(x, x’) < 2δ = ε;
- d(x’, y’) – d(x, y) ≤ d(y, y’) + d(x’, x) < 2δ = ε.
Hence, |d(x’, y’) – d(x, y)| ≤ ε. This shows d is continuous at (x, y). ♦
In addition, we had already proved that the standard arithmetic operations on real numbers are continuous. Thus, addition and product are continuous, as is reciprocal
Continuity in Topology
Since the notion of continuity can be described by open sets, let’s generalise it to topological spaces.
Definition. Let X and Y be topological spaces. A function f : X → Y on the underlying sets is said to be continuous at if:
- for any open subset containing f(a), there’s an open subset containing a.
In particular, f is continuous (at every point of X) if for any open , the pullback is open in X.
The following properties of continuity are obvious for any topological spaces X, Y and Z.
- The identity map id : X→ X is continuous.
- If f : X → Y is continuous at x and g : Y → Z is continuous at f(x), then gf : X → Z is continuous at x.
- If T and T’ are both topologies on X, then id : (X, T) → (X, T’) is continuous if and only if T is finer than T’.
From the third property, one sees that if f : X → Y is bijective and continuous, the inverse may not be continuous. Specifically, if T is strictly finer than T’, then the identity map (X, T) → (X, T’) is continuous but (X, T’) → (X, T) is not. Recall that if f : X → Y is bijective, continuous and has a continuous inverse, then we say f is a homeomorphism, in which case we get a bijection between the collection of open subsets of X and that of Y.
To check that a map is continuous, we don’t have to look at every open subset of Y.
Proposition. Let f : X→ Y be a map of topological spaces and be a subbasis of Y. Then f is continuous iff for any , is open.
The forward direction is obvious. The converse follows from the fact that the pullback preserves arbitrary intersection and union:
For example, to check that f : X → R is continuous, it only suffices to prove that and are open in X for any real a, b.
Next, we wish to prove that the standard maps are continuous.
- If is a subspace of a topological space, then the inclusion map is continuous.
- If X and Y are topological spaces, then the projection maps and are continuous.
- If is a collection of topological spaces and is the disjoint union, then every inclusion map is continuous.
For the first statement, each open pulls back to which is open in Y by definition of subspace. For the second statement, each open pulls back to which is open in X × Y by definition of product space. For the last statement, each open subset of X pulls back to which is open in Xi. ♦
In retrospect, one could even define topological spaces specifically to satisfy the above properties. And one defines the topology “just enough” such that the properties are satisfied. This will be expounded in a separate article.
Finally, we wish to show that continuity of a function depends “locally” on small neighbourhoods.
Theorem. If is a function of topological spaces and is a union of open subsets , then
- f is continuous if and only if is continuous for every i.
The forward direction is easy: since the inclusion map is continuous, so is the composition . Conversely, suppose each is continuous. Let V be an open subset of Y. Then
Now each is open in which is in turn open in X. Thus, is open in X. So , being a union of open subsets of X, is open in X. ♦
- Suppose f : X → Y is a continuous function of topological spaces and and are subspaces such that . Prove that the restriction is also continuous.
- Suppose f : X1 → Y1 and g: X2 → Y2 are continuous maps. Then the concatenation which takes , is also continuous.
- Let Y be a topological space written as a union of subspaces Prove that we get a continuous map which takes an element to the corresponding image in Y.
Answers (Highlight to read)
- Let V1 be an open subset of Y1. So we have V1 = V ∩ Y1 for some open subset V of Y. Then g-1(V1) = f-1(V) ∩ X1; by continuity of f, f-1(V) is open in X, so f-1(V) ∩ X1 is open in X1.
- Let U1 and U2 be open subsets of Y1 and Y2 respectively; thus U1 × U2 is a basic open subset of Y1 × Y2. Then h-1(U1 × U2) = f-1(U1) × g-1(U2) which is open in X1 × X2 since f and g are continuous. Thus the result follows.
- Let V be an open subset of Y. The inverse image in disjoint union of Xi is the disjoint union of Ui, where each Ui = Xi ∩ V is open in Xi by definition of subspace.