1. ## Proofs for Topology

Hey I don't know if anyone can help but here goes:I need to proof these 3 statements:

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

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.

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

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

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