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I am working on a proof where given a continuous surjection f:X$\to$Y, if X is connected then Y is connected. I am doing so by contraposition so I assumed Y was disconnected and am attempting to show X is disconnected.

By definition of disconnected, there exist open sets A' & B' in $\tau_Y$ such that A'$\cup$B'=Y and A'$\cap$B'=$\emptyset$. Since f is continuous, f$^{-1}$(A')=A & f$^{-1}$(B')=B are nonempty & in $\tau_X$. We wish to show A$\cup$B=X.

Since f(A$\cup$B)=f(A)$\cup$f(B) (by image of union) we have f(A$\cup$B)=A'$\cup$B' (by definition of A & B) which is equivalent to f(A$\cup$B)=Y and finally by surjection f(A$\cup$B)=f(X).

This is where I am stuck. How can I conclude that A$\cup$B=X?

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  • $\begingroup$ Are you stuck on proving $A\cup B\subseteq X$ or $X\subseteq A\cup B$? $\endgroup$ – bof Dec 11 '13 at 9:40
  • $\begingroup$ In order to show X is also disconnected I need to show A & B are disjoint (easy), open (easy), nonempty (easy), and that they span X (which is where I am stuck). To show that they span X I need to show that either A$\cup$B=X or that X$\subseteq$A$\cup$B. $\endgroup$ – Anonymous Dec 11 '13 at 9:43
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To show that $A\cup B=X$, we have to show that $A\cup B\subseteq X$ and that $X\subseteq A\cup B$.

To prove that $X\subseteq A\cup B$, we suppose $x\in X$ and prove that $x\in A\cup B$.

What does $x\in A\cup B$ mean? It means that $x\in A$ or $x\in B$,

which means that $x\in f^{-1}(A')$ or $x\in f^{-1}(B')$,

which means that $f(x)\in A'$ or $f(x)\in B'$,

which means that $f(x)\in A'\cup B'$,

which means that $f(x)\in Y$.

So the problem boils down to showing that $x\in X$ implies $f(x)\in Y$.

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  • $\begingroup$ I guess my problem here comes from when we have an x$\in$X that gives us an undefined f(x). Can we conclude that the undefined f(x)$\in$Y? For example a function that takes a to a, b to b, c to c where X={a,b,c,d} and Y={a,b,c}. Since d$\in$X we should be able to conclude f(d)$\in$Y but f(d) is undefined. $\endgroup$ – Anonymous Dec 11 '13 at 10:07
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Let $Y=A\cup B$ and $A\cap B=\emptyset$ where $A,B$ are open in $Y$. Then $f^{-1}(A),f^{-1}(B)$ are open in $X$ due to continuouity of $f$ and $f^{-1}(A)\cap f^{-1}(B)=\emptyset$ and hence $f^{-1}(A)=\emptyset $ or $f^{-1}(B)=\emptyset $ which means that $Y\cap A=\emptyset$ or $Y\cap B=\emptyset$. So $Y$ is connected.

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  • $\begingroup$ Our teacher never gave us this definition of connectedness. In order to use it in my proof I'd have to prove that it is logically equivalent to my current definition. $\endgroup$ – Anonymous Dec 11 '13 at 9:58
  • $\begingroup$ @Anonymous $Y$ is connected iff for open $A,B$ this is true $Y=A\cup B$ and $A\cap B=\emptyset$=>$Y\cap A=\emptyset$ or $Y\cap B=\emptyset$.See that the contraopposite of this is the definition of connectedness that you know. $\endgroup$ – Haha Dec 11 '13 at 10:18
  • $\begingroup$ Y is the space $\tau_Y$ is over. Necessarily, Y is nonempty. Therefore, your condition implies either A is empty or B is empty, or A or B is disjoint from Y. A and B are subsets of Y so they cannot be disjoint, and they cannot be empty as connected requires nonempty, disjoint, open sets whose union is the whole space. I don't see how those two are logically equivalent when one seems to contradict the other. $\endgroup$ – Anonymous Dec 11 '13 at 10:27
  • $\begingroup$ @Anonymous the one you know says when a set is not connedcted.mine says when a set is connected $\endgroup$ – Haha Dec 11 '13 at 10:52
  • $\begingroup$ That would explain why they are contradictory then. Sorry for the confusion! $\endgroup$ – Anonymous Dec 11 '13 at 11:04

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