Results 1 to 2 of 2

Thread: Proofs for Topology

  1. #1
    Newbie
    Joined
    Feb 2008
    Posts
    21

    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by reagan3nc View Post
    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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Order topology = discrete topology on a set?
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: Aug 6th 2011, 12:19 PM
  2. a topology such that closed sets form a topology
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: Jun 14th 2011, 05:43 AM
  3. Show quotient topology on [0,1] = usual topology on [0,1]
    Posted in the Differential Geometry Forum
    Replies: 4
    Last Post: Nov 5th 2010, 05:44 PM
  4. Replies: 0
    Last Post: May 4th 2009, 07:01 PM
  5. Topology of the Reals proofs
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Feb 12th 2009, 03:30 PM

Search Tags


/mathhelpforum @mathhelpforum