Proofs for Topology

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Originally Posted by reagan3nc

Let X have a discrete topolgy and Y be an arbitrary topological space. Show that every funtion f:X maps to Y is continuous.

Let $x_0 \in X$ we need to prove that for any $\epsilon > 0$ there exists $\delta > 0$ such that if $d(x,x_0)<\delta \mbox{ and }x\in X\implies d'(f(x),f(x_0))<\epsilon$ . Pick $\delta = 1/2$ then for $d(x,x_0)<\delta \mbox{ and }x\in X$ to be true we require that $x=x_0$ since it is the discrete metric. But then clearly $d'(f(x),f(x_0))=0<\epsilon$ .

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Let y have the trivial topology and X be an arbitrary topological space. Show that every funtion f: X map to Y is coninuous.

A function $f:X\mapsto Y$ from metric spaces $(X,d)$ and $(Y,d')$ is continous if and only if the inverse-image of every open set is open. Since $Y$ is a trivial topology it means the only open sets are $\emptyset$ and $Y$ . But $f^{-1} [\emptyset] = \emptyset$ and $f^{-1}[Y] = X$ . But $\emptyset$ and $X$ are open. Thus, $f$ was a continous function.

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Let X & Y be topological spaces. A function f:X maps to Y is continuous if and only in the inverse (C) is closed in X for every closed set C is a subset of Y.

Let $C$ be a closed subset of $Y$ . Consider $\bar C$ (the complement). This set is open. And since $f$ is continous it means $f^{-1}[\bar C] = \widehat{f^{-1}[C]}$ is open. Thus, $f^{-1}[C]$ is closed. The converse is similar.