Density is Continuous is L1 Norm Expanding Maps

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 $a\in X$ if:

for each ε>0, there exists δ>0 such that whenever $x\in X$ 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 $f(N_d(a, \delta))\subseteq N_{d'}(f(a), \epsilon)$ which is often more convenient.

If f is continuous at every a, we just say that f is continuous.

continuity_metric

Theorem. The function f : X → Y is continuous at a iff for any open subset $V\subseteq Y$ containing f(a), there's an open subset $U\subseteq f^{-1}(V)$ containing a.

In particular, f is continuous if and only if for any open subset $V\subseteq Y$ , the pull-back $f^{-1}(V)\subseteq X$ is open.

Proof.

Suppose f is continuous at a . If V is an open subset of Y containing f ( a ), then $N_{d'}(f(a),\epsilon)\subseteq V$ for some ε>0. By continuity at a , there exists δ>0 such that $f(N_d(a,\delta))\subseteq N_{d'}(f(a),\epsilon)$ . Hence $N_d(a,\delta) \subseteq f^{-1}(V)$ is an open subset containing a .

For the converse, suppose the stated condition holds. Given ε>0, $V := N_{d'}(f(a),\epsilon)$ is an open subset ofY containing f(a), so there is an open $U\subseteq f^{-1}(V)$ containinga. SinceU is open, there's an open ball $N(a,\delta)\subseteq U$ . It thus follows that $f(N(a,\delta))\subseteq V$ .

For the second statement, supposef is continuous and $V\subseteq Y$ is open. By what we just proved, each elementa of $U:= f^{-1}(V)$ is contained in an open subset $U_a \subseteq X$ which is contained inU. ThusU is a union of open subsets and is hence open.

Conversely, suppose whenever $V\subseteq Y$ is open, so is $U:=f^{-1}(V)\subseteq X.$ Leta be any point inX. From what we just proved,f is continuous ata. ♦

The theorem tells us that continuity is fundamentally a statement on the underlying topologies. In other words, iff : (X,d) → (Y,d') is a continuous map of metric spaces, then we can replaced ord' 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 $d:X\times X\to\mathbf{R}$ is continuous.

Proof

First, recall that there's no fixed way of defining a metric onX ×X, so we pick one of the many topologically equivalent ones, say $d_1((x,y), (x',y')) = d(x,x') + d(y,y').$

Let $(x,y)\in X\times X$ . To show continuity at that point, suppose ε>0. Letting δ = ε/2, we see that whenever $d_1((x, y), (x',y')) < \delta$ , we have:d(x,x') < δ andd(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 showsd is continuous at (x,y). ♦

Note

In addition, we had already proved that the standard arithmetic operations on real numbers are continuous. Thus, addition and product $\mathbf{R}\times \mathbf{R}\to \mathbf{R}$ are continuous, as is reciprocal $\mathbf{R}^*\to\mathbf{R}^*.$

blue-lin

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 becontinuous at $a\in X$ if:

In particular, f iscontinuous (at every point of X) if for any open $V\subseteq Y$ , the pullback $f^{-1}(V)$ is open in X.

The following properties of continuity are obvious for any topological spaces X,Y andZ.

The identity map id :X → X is continuous.
Iff :X →Y is continuous at x andg :Y →Z is continuous at f(x), then gf :X →Z is continuous atx.
IfT andT' are both topologies onX, then id : (X,T) → (X,T') is continuous if and only ifT is finer thanT'.

From the third property, one sees that iff :X →Y is bijective and continuous, the inverse $f^{-1}$ may not be continuous. Specifically, ifT is strictly finer thanT', then the identity map (X,T) → (X,T') is continuous but (X,T') → (X,T) is not. Recall that iff :X →Y is bijective, continuous and has a continuous inverse, then we sayf is ahomeomorphism, in which case we get a bijection between the collection of open subsets ofX and that ofY.

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 $S'\subseteq \mathbf{P}(Y)$ be a subbasis of Y. Then f is continuous iff for any $V\in S'$ , $f^{-1}(V) \subseteq X$ is open.

Proof.

The forward direction is obvious. The converse follows from the fact that the pullback preserves arbitrary intersection and union:

$f^{-1}(W_1 \cap W_2 \cap\ldots \cap W_k) = f^{-1}(W_1) \cap f^{-1}(W_2) \cap\ldots \cap f^{-1}(W_k),$

$f^{-1}(\cup V_i) = \cup_i f^{-1}(V_i).$ ♦

For example, to check thatf :X →R is continuous, it only suffices to prove that $f^{-1}((a, \infty))$ and $f^{-1}((-\infty, b))$ are open inX for any reala,b.

Next, we wish to prove that the standard maps are continuous.

Proposition.

Proof

For the first statement, each open $U\subseteq X$ pulls back to $i^{-1}(U) = U\cap Y$ which is open in Y by definition of subspace. For the second statement, each open $U\subseteq X$ pulls back to $p_1^{-1}(U) = U\times Y$ which is open inX ×Y by definition of product space. For the last statement, each open subset $U =\coprod_i U_i$ ofX pulls back to $\iota_i^{-1}(U) = U_i$ which is open inX_i . ♦

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 $f:X \to Y$ is a function of topological spaces and $X=\cup_i U_i$ is a union of open subsets $U_i \subseteq X$ , then

f is continuous if and only if $f|_{U_i} : U_i \to Y$ is continuous for every i.

Proof.

The forward direction is easy: since the inclusion map $\iota_i:U_i \hookrightarrow X$ is continuous, so is the composition $f\circ\iota_i = f|_{U_i}$ . Conversely, suppose each $f|_{U_i}$ is continuous. LetV be an open subset ofY. Then

$f^{-1}(V) = \cup_i (f^{-1}(V) \cap U_i) = \cup_i (f|_{U_i})^{-1}(V).$

Now each $(f|_{U_i})^{-1}(V)$ is open in $U_i$ which is in turn open inX. Thus, $(f|_{U_i})^{-1}(V)$ is open inX. So $f^{-1}(V)$ , being a union of open subsets of X, is open inX. ♦

Exercises

Supposef : X →Y is a continuous function of topological spaces and $X_1\subseteq X$ and $Y_1\subseteq Y$ are subspaces such that $f(X_1)\subseteq Y_1$ . Prove that the restriction $g := f|_{X_1} : X_1\to Y_1$ is also continuous.
Supposef :X ₁ →Y ₁ andg:X ₂ →Y ₂ are continuous maps. Then the concatenation $h:X_1\times X_2 \to Y_1\times Y_2,$ which takes $(x_1, x_2)\mapsto (f(x_1), g(x_2))$ , is also continuous.
LetY be a topological space written as a union of subspaces $Y = \cup_i X_i.$ Prove that we get a continuous map $\coprod_i X_i \to Y$ which takes an element $x_i \in X_i$ to the corresponding image inY.

Answers (Highlight to read)

Let V ₁ be an open subset ofY ₁. So we haveV ₁ =V ∩Y ₁ for some open subsetV ofY. Theng ^-1(V ₁) =f ^-1(V) ∩ X ₁; by continuity off,f ^-1(V) is open inX, so f ^-1(V) ∩X ₁ is open inX ₁.
LetU ₁ andU ₂ be open subsets ofY ₁ andY ₂ respectively; thusU ₁ ×U ₂ is a basic open subset ofY ₁ ×Y ₂. Thenh ^-1(U ₁ ×U ₂) =f ^-1(U ₁) × g ^-1(U ₂) which is open inX ₁ ×X ₂ sincef andg are continuous. Thus the result follows.
LetV be an open subset ofY. The inverse image in disjoint union ofX_i is the disjoint union of U_i , where each U_i =X_i ∩V is open in X_i by definition of subspace.

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Source: https://mathstrek.blog/2013/01/31/topology-continuous-maps/

Density is Continuous is L1 Norm Expanding Maps

Continuity in Metric Spaces

Continuity in Topology

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