24/10/2014

Metric Space: Continuity

                     Metric Space: Continuity






Definition

Let's recall the idea of continuity of functions. Continuity means, intuitively, that you can draw a function on a paper, without lifting your pen from it. Continuity is important in topology. But let's start in the beginning:
The classic delta-epsilon definition: Let (X,d),(Y,e) be spaces. A function f : X \rightarrow Y  is continuous at a point x if for all \epsilon_x > 0 there exists a \delta_{\epsilon_x} > 0such that: for all x_1 such that d(x,x_1) < \delta_{\epsilon_x} , we have that e(f(x), f(x_1)) < \epsilon_x.
Let's rephrase the definition to use balls: A function f : X \rightarrow Y  is continuous at a point x if for all \epsilon_x > 0 there exists \delta_{\epsilon_x} > 0 such that the following holds: for every x_1 such that x_1 \in B_{\delta_{\epsilon_x}} (x)  we have that  f(x_1) \in B_{\epsilon_{x}}(f(x)). Or more simply:  f(B_{\delta_{\epsilon_x}}(x)) \subseteq B_{\epsilon_{x}}(f(x))
Looks better already! But we can do more.
Definitions:
  • A function is continuous in a set S if it is continuous at every point in S.
  • A function is continuous if it is continuous in its entire domain.
Proposition: A function f : X \rightarrow Y  is continuous, by the definition above \Leftrightarrowfor every open set U in Y, The inverse image of Uf^{-1}(U), is open in X.
Note that f does not have to be surjective or bijective for f^{-1} to be well defined. The notation f^{-1} simply means f^{-1}(U) = \{x \in X: f(x) \in U\}.
Proof: First, let's assume that a function f is continuous by definition (The \Rightarrowdirection). We need to show that for every open set Uf^{-1}(U) is open.
Let U\subseteq Y be an open set. Let x \in f^{-1}(U)f(x) is in U and because U is open, we can find and \epsilon_x, such that B_{\epsilon_x}(f(x)) \subseteq U. Because f is continuous, for that \epsilon_x, we can find a \delta_{\epsilon_x} > 0 such that  f(B_{\delta_{\epsilon_x}}(x)) \subseteq B_{\epsilon_{x}}(f(x)) \subseteq U. that means that B_{\delta_{\epsilon_x}}(x) \subseteq f^{-1}(U), and therefore, x is an internal point. This is true for every x - meaning that all the points in f^{-1}(U) are internal points, and by definition, f^{-1}(U) is open.
(\Leftarrow)On the other hand, let's assume that for a function f for every open set U \in Yf^{-1}(U) is open in X. We need to show that f is continuous.
For every x\in X and for every \epsilon_x > 0, The set B_{\epsilon_x}(f(x)) is open in Y. Therefore the set V = f^{-1}(B_{\epsilon_x}(f(x))) is open in X. Note that x\in V. Because V is open, that means that we can find a \delta_{\epsilon_x} such that B_{\delta_{\epsilon_x}}(x) \subseteq V, and we have that  f(B_{\delta_{\epsilon_x}}(x)) \subseteq B_{\epsilon_{x}}(f(x)).