Given two distance spaces $D,E$ . If the function
$f:x\in D\rightarrow f(x)=y\in E$
is continuous, for an $x_n\rightarrow x\quad (n=1,2,\cdots)$
$f(x_n)\rightarrow f(x)\quad (n\rightarrow \infty)$ .
On the other hand, on topological spaces $S,T$,
if the function $f(S\rightarrow T)$ is continuous,
for an arbitrary open set $C_1\in T$ , $f^{-1}(C_1)$ becomes the open set
in $S$ .
This definition is equivalent to next two definitions.
For an arbitrary closed set $D_1\in T$ , if the function is continuous,
$f^{-1}(D_1)$ becomes the closed set in $S$ .
For $x\in S$ , let $f(x)=y\in T$ be. For the arbitrary neighbourhood $B_{\epsilon}(y)\in T$ ,
$f^{-1}(B_{\epsilon}(y))$ becomes the neighbourhood of $x$ in $S$ .
As no one will have the questions for the equality, the important thing is a proof.
2014/11/23
2014/11/12
topological spaces 4
A closed set is a complementary set of the open set.
Therefore, we are able to define the topology by closed sets.
As with open sets, a collection of closed sets $\mathcal{F}_c\in\Omega$
has following properties.
(1) If $C_1,C_2\in\mathcal{F}_c$ , then $(C_1\cup C_2)\in \mathcal{F}_c$ .
(2) If $\left\{C_i(i=1,2,\cdots)\right\}$ is a collection of the elements of $\mathcal{F}_c$ ,
then $\cap_{i=1}^{\infty}C_i\in\mathcal{F}_c$ .
These are gotten by properties of open sets. The topology by closed sets
will be defined by adding next requirements.
(3) $\Omega,\phi\in \mathcal{F}_c$ .
The proof of (2) is based on the famous De Morgan's laws.
$(\cap C_i)^c=\cup C_i^c$
Please try the challenge.
Therefore, we are able to define the topology by closed sets.
As with open sets, a collection of closed sets $\mathcal{F}_c\in\Omega$
has following properties.
(1) If $C_1,C_2\in\mathcal{F}_c$ , then $(C_1\cup C_2)\in \mathcal{F}_c$ .
(2) If $\left\{C_i(i=1,2,\cdots)\right\}$ is a collection of the elements of $\mathcal{F}_c$ ,
then $\cap_{i=1}^{\infty}C_i\in\mathcal{F}_c$ .
These are gotten by properties of open sets. The topology by closed sets
will be defined by adding next requirements.
(3) $\Omega,\phi\in \mathcal{F}_c$ .
The proof of (2) is based on the famous De Morgan's laws.
$(\cap C_i)^c=\cup C_i^c$
Please try the challenge.