Hi I'm learning real analysis that may interact some of the basic topology knowladge.
As far as I know that the formal definition of continues function is if $f(x)$ satisfy:
- $f:X\to Y$ is a function
- Define open-set for $<X,T_X><Y,T_Y>$ then $\forall S\in T_Y\rightarrow f^{-1}(S)\in T_X$ while $f^{-1}(S)=\left\{x\big|x\in X\land f(x)\in S\right\}$, that is the origin of an open set is an open set.
Thus if I have a sequence of function $\{f_i(x)\}$ that is continues, and they uniformly converge to $f(x)$ (so that I have to define distance as $<Y,d>$), can I prove that $f(x)$ is continues by the above definition??
If $(X, \tau ) $ is a topological space and $(Y, d)$ is a metric space and $f_n :X\to Y$ is a sequence of continuous functions then $f_n \to f $ uniformly if $v_n =\sup_{x\in X} d(f_n (x) , f(x))\to 0$ as $n\to \infty.$ Assume that $f_n \to f $ uniformly and $f_n$ are continuous functions for all $n$. Take any $x_0\in X$ and $\varepsilon >0. $ Then there exists $n_0 $ such that $$d (f_{n_0} (x) , f(x) )<\frac{\varepsilon }{3} $$ for all $x\in X.$ Since $f_{n_0}$ is continuous there exists an open neigborhood $U$ of $x_0 $ such that $$d(f_{n_0} (x_0 ) , f_{n_0 } (u) )<\frac{\varepsilon}{3} $$ for all $u\in U.$ Therefore $$d(f (x_0 ) , f(u) )\leq d(f (x_0 ) , f_{n_0 } (x_0) )+d(f_{n_0} (x_0 ) , f_{n_0 } (u) ) +d(f_{n_0} (u) , f (u) )<\varepsilon$$ hence $f$ is continuous at $x_0.$