Results 1 to 10 of 10

Thread: Continuous and connectivity

  1. #1
    Super Member Showcase_22's Avatar
    Joined
    Sep 2006
    From
    The raggedy edge.
    Posts
    782

    Continuous and connectivity

    Suppose that $\displaystyle \{A,B\}$ is a partition of a topological space $\displaystyle X$ and that $\displaystyle f: X \rightarrow Y$ is a map to another space $\displaystyle Y$. Prove that if the restrictions $\displaystyle f|A$ and $\displaystyle f|B$ are both continuous then $\displaystyle f$ is continuous.
    Let $\displaystyle E \subset X$ s.t $\displaystyle E \cap B \neq \phi$ and $\displaystyle E \cap A \neq \phi$.

    I need to show that $\displaystyle A$ and $\displaystyle B$ are connected (since if $\displaystyle f$ is continuous on $\displaystyle A$ and $\displaystyle B$, then the only place it might be discontinuous is at the boundary, where $\displaystyle A$ meets $\displaystyle B$).

    $\displaystyle A$ is connected (as is $\displaystyle B$) since $\displaystyle f$ is a continuous map between any two points.

    This is what I know, but I can't see what I can do from here! Can someone help?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member Tinyboss's Avatar
    Joined
    Jul 2008
    Posts
    433
    What's the definition of a partition? I'm going to assume it's not just "sets whose union is the space", else a counterexample is $\displaystyle X=[0,2],\:A=[0,1),\:B=[1,2]$ and

    $\displaystyle f(x)=\begin{cases}0&x\in A\\1&x\in B.\end{cases}$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member Showcase_22's Avatar
    Joined
    Sep 2006
    From
    The raggedy edge.
    Posts
    782
    If $\displaystyle \{A,B\}$ is a partition of a set $\displaystyle X$, then $\displaystyle A \cup B=X$ and $\displaystyle A \cap B= \phi$.

    I didn't think the second condition helped, so I left it out.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member Tinyboss's Avatar
    Joined
    Jul 2008
    Posts
    433
    Are you sure there's not another condition, such as A and B are both open or both closed? Else it can fail, as I demonstrated above.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member Showcase_22's Avatar
    Joined
    Sep 2006
    From
    The raggedy edge.
    Posts
    782
    Ah yes! Both $\displaystyle A,B$ have to be open.

    Sorry! I forgot the most important condition.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member Tinyboss's Avatar
    Joined
    Jul 2008
    Posts
    433
    Quote Originally Posted by Showcase_22 View Post
    Ah yes! Both $\displaystyle A,B$ have to be open.

    Sorry! I forgot the most important condition.
    Well, there's a "lemma" that talks about "pasting" continuous functions together. ;-) I don't know if it would be kosher just to cite it, but the proof isn't hard.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member Showcase_22's Avatar
    Joined
    Sep 2006
    From
    The raggedy edge.
    Posts
    782
    Okay.

    I know that A and B are open. Since f is continuous then $\displaystyle f(A), \ f(B)$ are also open.

    $\displaystyle Im(f)=f(A) \cup f(B)$. This is the union of two open sets so $\displaystyle Im(f)$ is open.

    Therefore for any $\displaystyle E$ as defined before, $\displaystyle f(E)$ is also going to be open so $\displaystyle f$ maps an open set to an open set. Therefore f is continuous.

    I think that's about right. The odd thing is that this is in the connectedness chapter, that's why I was trying to approach it from a connectedness angle.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Senior Member Tinyboss's Avatar
    Joined
    Jul 2008
    Posts
    433
    Quote Originally Posted by Showcase_22 View Post
    I know that A and B are open. Since f is continuous then $\displaystyle f(A), \ f(B)$ are also open.
    This is false--it only works in the inverse direction. For example, constant functions are continuous, but their images are single points. And the image of $\displaystyle f(x)=x^2$ is $\displaystyle [0,\infty)$, also not open.

    A function with the property you gave is called an "open map".
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Super Member Showcase_22's Avatar
    Joined
    Sep 2006
    From
    The raggedy edge.
    Posts
    782
    Sorry I keep getting this wrong! I think i'm getting either flustered or tired.

    let $\displaystyle V \in Y$ be open.

    Then $\displaystyle f^{-1}(V) \subset A \cup B$.

    If $\displaystyle f^{-1}(V) \subset A$ or $\displaystyle f^{-1}(V) \subset B$ then $\displaystyle f^{-1}(V)$ is open since we know that $\displaystyle f$ is continuous on these intervals.

    If $\displaystyle f^{-1}(V)$ has points in both $\displaystyle A$ and $\displaystyle B$ then partition $\displaystyle V$ into two sets $\displaystyle I,J$ where $\displaystyle f^{-1}(I) \subset A$ and $\displaystyle f^{-1}(J) \subset B$.

    $\displaystyle f^{-1}(I),\ f^{-1}(J)$ are both open since $\displaystyle f$ is continuous on both $\displaystyle A$ and $\displaystyle B$.

    Hence $\displaystyle f^{-1}(V)$ is always open. The preimage of an open set is also an open set so $\displaystyle f$ is continuous.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Member Focus's Avatar
    Joined
    Aug 2009
    Posts
    228
    Quote Originally Posted by Showcase_22 View Post
    Sorry I keep getting this wrong! I think i'm getting either flustered or tired.

    let $\displaystyle V \in Y$ be open.

    Then $\displaystyle f^{-1}(V) \subset A \cup B$.

    If $\displaystyle f^{-1}(V) \subset A$ or $\displaystyle f^{-1}(V) \subset B$ then $\displaystyle f^{-1}(V)$ is open since we know that $\displaystyle f$ is continuous on these intervals.

    If $\displaystyle f^{-1}(V)$ has points in both $\displaystyle A$ and $\displaystyle B$ then partition $\displaystyle V$ into two sets $\displaystyle I,J$ where $\displaystyle f^{-1}(I) \subset A$ and $\displaystyle f^{-1}(J) \subset B$.

    $\displaystyle f^{-1}(I),\ f^{-1}(J)$ are both open since $\displaystyle f$ is continuous on both $\displaystyle A$ and $\displaystyle B$.

    Hence $\displaystyle f^{-1}(V)$ is always open. The preimage of an open set is also an open set so $\displaystyle f$ is continuous.
    How do you know that both I and J are open?

    Here is a simple way to do this, take U open in Y, then $\displaystyle f^{-1}|_A(U)=f^{-1}(U) \cap A,f^{-1}|_B(U)=f^{-1}(U) \cap B$ are both open in X. Then $\displaystyle f^{-1}(U)=f^{-1}(U) \cap (A\cup B)=(f^{-1}(U) \cap A)\cup(f^{-1}(U) \cap B)$ a union of two open sets, which is open.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Connectivity Proof
    Posted in the Discrete Math Forum
    Replies: 0
    Last Post: Oct 19th 2010, 02:58 PM
  2. Connectivity Graphs
    Posted in the Discrete Math Forum
    Replies: 0
    Last Post: Apr 30th 2010, 02:34 AM
  3. Connectivity
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: Feb 1st 2010, 04:18 AM
  4. connectivity question in R^{n}
    Posted in the Differential Geometry Forum
    Replies: 9
    Last Post: Jan 5th 2010, 05:33 PM
  5. Edge Connectivity
    Posted in the Discrete Math Forum
    Replies: 0
    Last Post: May 29th 2008, 06:08 AM

Search Tags


/mathhelpforum @mathhelpforum