3.4 KiB
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)\coloneq \{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: f_n:X\to Y\in Z continuous,
Which is to prove the uniform convergence,
f_n \to f \in \operatorname{Map}(X,Y)
Then we want to show that f is continuous.
Let B_r(y) be an arbitrary ball in Y, it suffices to show that f^{-1}(B_r(y)) is open in X.
Take N to be large enough such that for n\geq N, we have
\rho(f_n(x), f(x)) < \frac{r}{3}
In particular, this holds for n=N. So we have
d(f_N(x), f(x)) < \frac{r}{3},\forall x\in X
Take x_0\in f^{-1}(B_r(y)), we'd like to show that there is an open ball around x_0 in f^{-1}(B_r(y)).
Since f_N is continuous, f^{-1}_N(B_{\frac{r}{3}}(y)) is open in X.
d(f(x_0), f(x_0))<\frac{r}{3}
continue the proof in bonus video