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.
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