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Math4201 Lecture 16 (Topology I)
Continuous maps
The following maps are continuous:
F_+:\mathbb{R}\times \mathbb{R}\to \mathbb{R}, (x,y)\to x+y
F_-:\mathbb{R}\times \mathbb{R}\to \mathbb{R}, (x,y)\to x-y
F_\times:\mathbb{R}\times \mathbb{R}\to \mathbb{R}, (x,y)\to x\times y
F_\div:\mathbb{R}\times (\mathbb{R}\setminus \{0\})\to \mathbb{R}, (x,y)\to \frac{x}{y}
Composition of continuous functions is continuous
Let f,g:X\to \mathbb{R} be continuous functions. X is topological space.
Then the following functions are continuous:
H:X\to \mathbb{R}\times \mathbb{R}, x\to (f(x),g(x))
Since the composition of continuous functions is continuous, we have
F_+\circ H:X\to \mathbb{R}, x\to f(x)+g(x)
F_-\circ H:X\to \mathbb{R}, x\to f(x)-g(x)
F_\times\circ H:X\to \mathbb{R}, x\to f(x)\times g(x)
are all continuous.
More over, if g(x)\neq 0 for all x\in X, then
F_\div\circ H:X\to \mathbb{R}, x\to \frac{f(x)}{g(x)}
is continuous following the similar argument.
Defining metric for functions
Definition of bounded metric space
A metric space (Y,d) is bounded if there is M\in\mathbb{R}^{\geq 0} such that
\forall y,y'\in Y, d(y,y')<M
Example of bounded metric space
If (Y,d) is a bounded metric space, let M be a positive constant, then \overline{d}=\min\{M,d\} is a bounded metric space.
In fact, the metric topology by d and \overline{d} are the same. (proved in homeworks)
Let X be a topological space. and (Y,d) be a bounded metric space.
\operatorname{Map}(X,Y)\coloneqq \{f:X\to Y|f \text{ is a map}\}
Define \rho:\operatorname{Map}(X,Y)\times \operatorname{Map}(X,Y)\to \mathbb{R} by
\rho(f,g)=\sup_{x\in X} d(f(x),g(x))
Lemma space of map with metric defined is a metric space
(\operatorname{Map}(X,Y),\rho) is a metric space.
Proof
Proof is similar to showing that the square metric is a metric on \mathbb{R}^n.
\rho(f,g)=0\implies \sup_{x\in X}(d(f(x),g(x)))=0
Since d(f(x),g(x))\geq 0, this implies that d(f(x),g(x))=0 for all x\in X.
The triangle inequality of being metric for \rho follows from the similar properties for d.
Lemma continuous maps form a closed subset of the space of maps
Let (\operatorname{Map}(X,Y),\rho) be a metric space defined before.
and
Z=\{f:X\to Y|f \text{ is a continuous map}\}
Then Z is a closed subset of (\operatorname{Map}(X,Y),\rho).
Proof
We need to show that \overline{Z}=Z.
Since \operatorname{Map}(X,Y) is a metric space, this is equivalent to showing that: Let f_n:X\to Y\in Z be a sequence of continuous maps,
Which is to prove the uniform convergence,
f_n \to f \in \operatorname{Map}(X,Y)
Then we want to show that f is also continuous.
It is to show that for any open subspace V of Y, f^{-1}(V) is open in X.
Take x_0\in f^{-1}(V), we'd like to show that there is an open neighborhood U of x_0 such that U\subseteq f^{-1}(V).
Since x_0\in f^{-1}(V), then f(x_0)\in V. By metric definition, there is r>0 such that B_r(f(x_0))\subseteq V.
Take N to be large enough such \rho(f_N(x), f(x)) < \frac{r}{3}
So \forall x\in X, d(f(x),f_N(x))<\frac{r}{3}
Since f_N is continuous, f_N^{-1}(B_{r/3}(f(x_0))) is an open set U\subseteq X containing x_0.
Take x\in U, d(f(x),f(x_0))<d(f(x),f_N(x_0))+d(f_N(x),f_N(x_0))+d(f_N(x_0),f(x_0)) using triangle inequality.
Note that,
d(f(x),f_N(x))<\frac{r}{3} (using N large enough),
d(f_N(x),f_N(x_0))<\frac{r}{3} (using x\in U, then f_N(x)\in B_{r/3}(f_N(x_0)), so d(f_N(x),f_N(x_0))<\frac{r}{3}),
d(f_N(x_0),f(x_0))<\frac{r}{3} (using N large enough),
So d(f(x),f(x_0))<\frac{r}{3}+\frac{r}{3}+\frac{r}{3}=r.
So f(x)\in B_r(f(x_0))\implies x\in f^{-1}(B_r(f(x_0)))\implies x\in f^{-1}(V)\implies U\subseteq f^{-1}(V).
So f^{-1}(V) is open in X.