Continuous function maps connected set onto connected set

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Let $X, Y$ be topological spaces and $f: Xto Y$ continuous. Show that $f$ maps connected sets onto connected sets.




I am not sure if I understand the definition of a connected set right.
Our definitions are as follows:



A topological space $X$ is "connected" if $emptyset$ and $X$ are the only sets, which are closed and open in $X$.



A subset of a topological space is "connected" if it is "connected" with regards to the subspace topology.



Therefor a set $Asubseteq X$ is connected, if $A$ and $emptyset$ are the only closed and open ("clopen") sets with regards to the subspace topology?



Now for the proof:



Let $emptysetneq Asubset X$ be connected. Therefor $A$ is not clopen with regards to the subspace topology.



I have to show, that $f(A)$ is not clopen with regards to the subspace topology.



Suppose $emptysetneq f(A)subset Y$ is clopen. Since $f$ is continuous the preimage of open and closed sets is open and closed. Hence $A$ is clopen. Which contradicts the assumption.



Is this correct?
Thanks in advance.







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  • In your proof, you first assume that $A$ is a connected subset of $X$. It looks like you will try to prove that $f(A)$ is connected. Then conclude that $A$ is not clopen with respect to the subspace topology. But $A$ as a subspace is clopen. You say that your aim is to prove that $f(A)$ is not clopen with respect to the subspace topology, but again it is.
    – user583185
    Aug 13 at 10:57










  • So, I take $emptysetneq Bsubset A$ which is not clopen, since $A$ is connected and I have to show, that $f(B)$ is not clopen?
    – Cornman
    Aug 13 at 10:57










  • @rTSlines My comment above is refering to your first comment. Was it wrong?
    – Cornman
    Aug 13 at 10:58










  • Well, to begin with it is not necessary to prove that $f$ maps all connected subsets of $X$ to connected. It is enough to assume that $X$ is connected and prove that $f(X)$ is connected. Since $X$ and $Y$ are arbitrary, with $X$ any connected space, then you are proving the required statement.
    – user583185
    Aug 13 at 10:58











  • So, what you need to show is that no proper subset of $f(X)$ is clopen. Assume $Bsubset f(X)$ is clopen. Then look at $f^-1(B)subset X$.
    – user583185
    Aug 13 at 11:00















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0
down vote

favorite













Let $X, Y$ be topological spaces and $f: Xto Y$ continuous. Show that $f$ maps connected sets onto connected sets.




I am not sure if I understand the definition of a connected set right.
Our definitions are as follows:



A topological space $X$ is "connected" if $emptyset$ and $X$ are the only sets, which are closed and open in $X$.



A subset of a topological space is "connected" if it is "connected" with regards to the subspace topology.



Therefor a set $Asubseteq X$ is connected, if $A$ and $emptyset$ are the only closed and open ("clopen") sets with regards to the subspace topology?



Now for the proof:



Let $emptysetneq Asubset X$ be connected. Therefor $A$ is not clopen with regards to the subspace topology.



I have to show, that $f(A)$ is not clopen with regards to the subspace topology.



Suppose $emptysetneq f(A)subset Y$ is clopen. Since $f$ is continuous the preimage of open and closed sets is open and closed. Hence $A$ is clopen. Which contradicts the assumption.



Is this correct?
Thanks in advance.







share|cite|improve this question






















  • In your proof, you first assume that $A$ is a connected subset of $X$. It looks like you will try to prove that $f(A)$ is connected. Then conclude that $A$ is not clopen with respect to the subspace topology. But $A$ as a subspace is clopen. You say that your aim is to prove that $f(A)$ is not clopen with respect to the subspace topology, but again it is.
    – user583185
    Aug 13 at 10:57










  • So, I take $emptysetneq Bsubset A$ which is not clopen, since $A$ is connected and I have to show, that $f(B)$ is not clopen?
    – Cornman
    Aug 13 at 10:57










  • @rTSlines My comment above is refering to your first comment. Was it wrong?
    – Cornman
    Aug 13 at 10:58










  • Well, to begin with it is not necessary to prove that $f$ maps all connected subsets of $X$ to connected. It is enough to assume that $X$ is connected and prove that $f(X)$ is connected. Since $X$ and $Y$ are arbitrary, with $X$ any connected space, then you are proving the required statement.
    – user583185
    Aug 13 at 10:58











  • So, what you need to show is that no proper subset of $f(X)$ is clopen. Assume $Bsubset f(X)$ is clopen. Then look at $f^-1(B)subset X$.
    – user583185
    Aug 13 at 11:00













up vote
0
down vote

favorite









up vote
0
down vote

favorite












Let $X, Y$ be topological spaces and $f: Xto Y$ continuous. Show that $f$ maps connected sets onto connected sets.




I am not sure if I understand the definition of a connected set right.
Our definitions are as follows:



A topological space $X$ is "connected" if $emptyset$ and $X$ are the only sets, which are closed and open in $X$.



A subset of a topological space is "connected" if it is "connected" with regards to the subspace topology.



Therefor a set $Asubseteq X$ is connected, if $A$ and $emptyset$ are the only closed and open ("clopen") sets with regards to the subspace topology?



Now for the proof:



Let $emptysetneq Asubset X$ be connected. Therefor $A$ is not clopen with regards to the subspace topology.



I have to show, that $f(A)$ is not clopen with regards to the subspace topology.



Suppose $emptysetneq f(A)subset Y$ is clopen. Since $f$ is continuous the preimage of open and closed sets is open and closed. Hence $A$ is clopen. Which contradicts the assumption.



Is this correct?
Thanks in advance.







share|cite|improve this question















Let $X, Y$ be topological spaces and $f: Xto Y$ continuous. Show that $f$ maps connected sets onto connected sets.




I am not sure if I understand the definition of a connected set right.
Our definitions are as follows:



A topological space $X$ is "connected" if $emptyset$ and $X$ are the only sets, which are closed and open in $X$.



A subset of a topological space is "connected" if it is "connected" with regards to the subspace topology.



Therefor a set $Asubseteq X$ is connected, if $A$ and $emptyset$ are the only closed and open ("clopen") sets with regards to the subspace topology?



Now for the proof:



Let $emptysetneq Asubset X$ be connected. Therefor $A$ is not clopen with regards to the subspace topology.



I have to show, that $f(A)$ is not clopen with regards to the subspace topology.



Suppose $emptysetneq f(A)subset Y$ is clopen. Since $f$ is continuous the preimage of open and closed sets is open and closed. Hence $A$ is clopen. Which contradicts the assumption.



Is this correct?
Thanks in advance.









share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 13 at 10:49

























asked Aug 13 at 10:38









Cornman

2,61921128




2,61921128











  • In your proof, you first assume that $A$ is a connected subset of $X$. It looks like you will try to prove that $f(A)$ is connected. Then conclude that $A$ is not clopen with respect to the subspace topology. But $A$ as a subspace is clopen. You say that your aim is to prove that $f(A)$ is not clopen with respect to the subspace topology, but again it is.
    – user583185
    Aug 13 at 10:57










  • So, I take $emptysetneq Bsubset A$ which is not clopen, since $A$ is connected and I have to show, that $f(B)$ is not clopen?
    – Cornman
    Aug 13 at 10:57










  • @rTSlines My comment above is refering to your first comment. Was it wrong?
    – Cornman
    Aug 13 at 10:58










  • Well, to begin with it is not necessary to prove that $f$ maps all connected subsets of $X$ to connected. It is enough to assume that $X$ is connected and prove that $f(X)$ is connected. Since $X$ and $Y$ are arbitrary, with $X$ any connected space, then you are proving the required statement.
    – user583185
    Aug 13 at 10:58











  • So, what you need to show is that no proper subset of $f(X)$ is clopen. Assume $Bsubset f(X)$ is clopen. Then look at $f^-1(B)subset X$.
    – user583185
    Aug 13 at 11:00

















  • In your proof, you first assume that $A$ is a connected subset of $X$. It looks like you will try to prove that $f(A)$ is connected. Then conclude that $A$ is not clopen with respect to the subspace topology. But $A$ as a subspace is clopen. You say that your aim is to prove that $f(A)$ is not clopen with respect to the subspace topology, but again it is.
    – user583185
    Aug 13 at 10:57










  • So, I take $emptysetneq Bsubset A$ which is not clopen, since $A$ is connected and I have to show, that $f(B)$ is not clopen?
    – Cornman
    Aug 13 at 10:57










  • @rTSlines My comment above is refering to your first comment. Was it wrong?
    – Cornman
    Aug 13 at 10:58










  • Well, to begin with it is not necessary to prove that $f$ maps all connected subsets of $X$ to connected. It is enough to assume that $X$ is connected and prove that $f(X)$ is connected. Since $X$ and $Y$ are arbitrary, with $X$ any connected space, then you are proving the required statement.
    – user583185
    Aug 13 at 10:58











  • So, what you need to show is that no proper subset of $f(X)$ is clopen. Assume $Bsubset f(X)$ is clopen. Then look at $f^-1(B)subset X$.
    – user583185
    Aug 13 at 11:00
















In your proof, you first assume that $A$ is a connected subset of $X$. It looks like you will try to prove that $f(A)$ is connected. Then conclude that $A$ is not clopen with respect to the subspace topology. But $A$ as a subspace is clopen. You say that your aim is to prove that $f(A)$ is not clopen with respect to the subspace topology, but again it is.
– user583185
Aug 13 at 10:57




In your proof, you first assume that $A$ is a connected subset of $X$. It looks like you will try to prove that $f(A)$ is connected. Then conclude that $A$ is not clopen with respect to the subspace topology. But $A$ as a subspace is clopen. You say that your aim is to prove that $f(A)$ is not clopen with respect to the subspace topology, but again it is.
– user583185
Aug 13 at 10:57












So, I take $emptysetneq Bsubset A$ which is not clopen, since $A$ is connected and I have to show, that $f(B)$ is not clopen?
– Cornman
Aug 13 at 10:57




So, I take $emptysetneq Bsubset A$ which is not clopen, since $A$ is connected and I have to show, that $f(B)$ is not clopen?
– Cornman
Aug 13 at 10:57












@rTSlines My comment above is refering to your first comment. Was it wrong?
– Cornman
Aug 13 at 10:58




@rTSlines My comment above is refering to your first comment. Was it wrong?
– Cornman
Aug 13 at 10:58












Well, to begin with it is not necessary to prove that $f$ maps all connected subsets of $X$ to connected. It is enough to assume that $X$ is connected and prove that $f(X)$ is connected. Since $X$ and $Y$ are arbitrary, with $X$ any connected space, then you are proving the required statement.
– user583185
Aug 13 at 10:58





Well, to begin with it is not necessary to prove that $f$ maps all connected subsets of $X$ to connected. It is enough to assume that $X$ is connected and prove that $f(X)$ is connected. Since $X$ and $Y$ are arbitrary, with $X$ any connected space, then you are proving the required statement.
– user583185
Aug 13 at 10:58













So, what you need to show is that no proper subset of $f(X)$ is clopen. Assume $Bsubset f(X)$ is clopen. Then look at $f^-1(B)subset X$.
– user583185
Aug 13 at 11:00





So, what you need to show is that no proper subset of $f(X)$ is clopen. Assume $Bsubset f(X)$ is clopen. Then look at $f^-1(B)subset X$.
– user583185
Aug 13 at 11:00











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You are saying that $Asubseteq X$ is not clopen wrt subspace topology, but this statement is not true. Every subset $A$ of $X$ is clopen wrt the subspace topology on $A$ inherited from $X$.



Under the condition that $A$ is connected it must be proved that $f(A)$ is connected (not that $f(A)$ is not clopen).



This comes to the same as proving that no set $Usubset f(A)$ exists that is clopen wrt the subspace topology of $f(A)$ and satisfies $Unotinvarnothing,f(A)$.



Suppose that there is such a set.



Then its preimage under $f$ is a subset $V$ of $A$ that is clopen wrt the subspace topology of $A$ and satisfies $Vnotinvarnothing, A$.



Proved is then: $f(A)$ not connected $implies A$ not connected, or equivalently:



$$Atext connected implies f(A)text connected$$






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    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    1
    down vote



    accepted










    You are saying that $Asubseteq X$ is not clopen wrt subspace topology, but this statement is not true. Every subset $A$ of $X$ is clopen wrt the subspace topology on $A$ inherited from $X$.



    Under the condition that $A$ is connected it must be proved that $f(A)$ is connected (not that $f(A)$ is not clopen).



    This comes to the same as proving that no set $Usubset f(A)$ exists that is clopen wrt the subspace topology of $f(A)$ and satisfies $Unotinvarnothing,f(A)$.



    Suppose that there is such a set.



    Then its preimage under $f$ is a subset $V$ of $A$ that is clopen wrt the subspace topology of $A$ and satisfies $Vnotinvarnothing, A$.



    Proved is then: $f(A)$ not connected $implies A$ not connected, or equivalently:



    $$Atext connected implies f(A)text connected$$






    share|cite|improve this answer
























      up vote
      1
      down vote



      accepted










      You are saying that $Asubseteq X$ is not clopen wrt subspace topology, but this statement is not true. Every subset $A$ of $X$ is clopen wrt the subspace topology on $A$ inherited from $X$.



      Under the condition that $A$ is connected it must be proved that $f(A)$ is connected (not that $f(A)$ is not clopen).



      This comes to the same as proving that no set $Usubset f(A)$ exists that is clopen wrt the subspace topology of $f(A)$ and satisfies $Unotinvarnothing,f(A)$.



      Suppose that there is such a set.



      Then its preimage under $f$ is a subset $V$ of $A$ that is clopen wrt the subspace topology of $A$ and satisfies $Vnotinvarnothing, A$.



      Proved is then: $f(A)$ not connected $implies A$ not connected, or equivalently:



      $$Atext connected implies f(A)text connected$$






      share|cite|improve this answer






















        up vote
        1
        down vote



        accepted







        up vote
        1
        down vote



        accepted






        You are saying that $Asubseteq X$ is not clopen wrt subspace topology, but this statement is not true. Every subset $A$ of $X$ is clopen wrt the subspace topology on $A$ inherited from $X$.



        Under the condition that $A$ is connected it must be proved that $f(A)$ is connected (not that $f(A)$ is not clopen).



        This comes to the same as proving that no set $Usubset f(A)$ exists that is clopen wrt the subspace topology of $f(A)$ and satisfies $Unotinvarnothing,f(A)$.



        Suppose that there is such a set.



        Then its preimage under $f$ is a subset $V$ of $A$ that is clopen wrt the subspace topology of $A$ and satisfies $Vnotinvarnothing, A$.



        Proved is then: $f(A)$ not connected $implies A$ not connected, or equivalently:



        $$Atext connected implies f(A)text connected$$






        share|cite|improve this answer












        You are saying that $Asubseteq X$ is not clopen wrt subspace topology, but this statement is not true. Every subset $A$ of $X$ is clopen wrt the subspace topology on $A$ inherited from $X$.



        Under the condition that $A$ is connected it must be proved that $f(A)$ is connected (not that $f(A)$ is not clopen).



        This comes to the same as proving that no set $Usubset f(A)$ exists that is clopen wrt the subspace topology of $f(A)$ and satisfies $Unotinvarnothing,f(A)$.



        Suppose that there is such a set.



        Then its preimage under $f$ is a subset $V$ of $A$ that is clopen wrt the subspace topology of $A$ and satisfies $Vnotinvarnothing, A$.



        Proved is then: $f(A)$ not connected $implies A$ not connected, or equivalently:



        $$Atext connected implies f(A)text connected$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Aug 13 at 10:58









        drhab

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