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I'm trying to understand the proof of:

If $f: X \rightarrow Y$ is continuous, and $X$ is connected, then $f(X)$ is connected.

What are we trying to do in the following proof (are we proving the contrapositive or trying to prove by contradiction)?

Suppose that $f(X) = A \cup B$ is a separation. Let $C = f^{-1}(A), D = f^{-1}(B)$. Then $C,D \neq \varnothing, C \cap D = \varnothing$. Write $A = f(X) \cap U$, where $U \subset Y$ is open. Then $C = f^{-1}(A)=f^{-1}(U)$ is open since $f$ is continuous. Similarly, $D$ is open.

Irregular User
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  • You can think of it as proving the contrapositive or as proof by contradiction. There's not really a difference. – Max May 29 '16 at 02:31
  • Yes, I'm not too sure either and would like to know. Apparently the trick of the proof is to realise that $f^{-1}(A) = f^{-1}(U)$. – Irregular User May 29 '16 at 02:42
  • Where did you find the proof? –  May 29 '16 at 02:44
  • @Jack They come from notes from a lecture. – Irregular User May 29 '16 at 02:44
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    @Max There is a difference between contrapositive and contradiction. –  May 29 '16 at 02:55
  • You may also use the theorem that, "A topological space $X$ is connected iff every continuous function $f:X\to {\pm 1}$ constant". This gives an elegant and beautiful proof. –  May 29 '16 at 04:07
  • @user170039 That's the first time I've heard of that theorem and it's not in my notes, could you possibly link to it? – Irregular User May 29 '16 at 04:11
  • Let me just give a short outline to it's proof, I hope you can fill out the details. Suppose that $X$ be a connected space and $f:X\to {\pm 1}$ is a continuous function. We are to show that $f$ is constant. If it's not then $f$ is onto. Hence both $f^{-1}(-1)$ and $f^{-1}(1)$ are non-empty open subsets of $X$. Observe that - (1) they are both closed and (2) they are disjoint. This shows, $X=f^{-1}(-1)\cup f^{-1}(1)$, a contradiction. Can you try the converse part? –  May 29 '16 at 04:19
  • You can also take help from this. –  May 29 '16 at 05:07

2 Answers2

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Suppose that $f(X) = A \cup B$ is a separation.

It is assumed here that $f(X)$ is not connected so that we can reach some contradiction.

Let $C = f^{-1}(A), D = f^{-1}(B)$. Then $C,D \neq \varnothing, C \cap D = \varnothing$. $\tag{1}$

This comes from the assumption that $A\cup B$ is a separation. [Added:]Note also that in this step, we also have $$ X=C\cup D. $$ Thus if we can prove that both $C$ and $D$ are open (in $X$), then we will have a desired contradiction.

Write $A = f(X) \cap U$, where $U \subset Y$ is open. Then $C = f^{-1}(A)=f^{-1}(U)$ is open since $f$ is continuous. Similarly, $D$ is open.

In this step, we are trying to argue that both $C$ and $D$ are open. Thus this fact (both $C$ and $D$ being open) and (1) together imply that $X$ is not connected, which is a desired contradiction.

[Added:] The reason we can write $A$ as the proof puts is that $A$ is assumed to be open in $f(X)$ with the subspace topology.

Let $P$ be that "$f:X\to Y$ is a continuous map between two topological spaces $X$ and $Y$". Let $Q$ be that "$X$ is a connected topological space". Let $R$ be that $f(X)$ is connected. So you statement is

$P$ and $Q$ implies $R$. (Or, if P and Q, then R.)

What we are trying to do in the proof in your question is that assuming $R$ is not true and $P$ is true, prove that $Q$ is not true.

The difference between the Contrapositive method and the Contradiction method is subtle. Let's examine how the two methods work when trying to prove "If P, Then Q".

  • Method of Contradiction: Assume P and Not Q and prove some sort of contradiction.
  • Method of Contrapositive: Assume Not Q and prove Not P.

The method of Contrapositive has the advantage that your goal is clear: Prove Not P. In the method of Contradiction, your goal is to prove a contradiction, but it is not always clear what the contradiction is going to be at the start.

See more for this in the question Proof by contradiction vs Prove the contrapositive.

Community
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  • One should have $U\cap V\supset f(X)$ instead of $U\cap V=f(X)$. –  May 29 '16 at 03:45
  • You don't need those $U$ and $V$ to show $X=C\cup D$. –  May 29 '16 at 03:49
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    Wonderful, thanks for your efforts. I genuinely feel like I've finally understood this proof. I feel like it was a lot more work than it should have been since really, this proof isn't difficult, but understanding why certain elements were there was. – Irregular User May 29 '16 at 03:52
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The common definition of connectedness in topology is that $X$ is connected if it is not disconnected, where disconnected means that $X$ is a non-trivial union of two disjoint open sets. Thus, in a sense, connectedness is defined as the absence of disconnectedness (in contract to the definition of path connectedness, which is more intuitive in that it assert the existence of a connecting entity between any two points). This is possibly the source of your confusion. So, think of restating the theorem in terms of disconnectedness: If $f\colon X\to Y$ is continuous then $f(X)$ being disconnected implies $X$ must be disconnected. Convince yourself that this is equivalent your statement of the result, and then that the proof is directly showing precisely that.

Comment: It is possible to define connectedness directly, by means of connecting entities between any two points. See here for details, and in particular a proof of the result you mention that does not use contrapositive or proof by contradiction.

Ittay Weiss
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  • Thanks for the explanation, but I'm already familiar with the definition of connectedness. I'm just unclear with why some things in the proof are done. For example, why we've defined $A$ in the way we have. – Irregular User May 29 '16 at 02:42
  • Well, think of it this way: you assume a separation of $f(X)$, i.e., open sets $A,B$ that demonstrate $f(X)$ is disconnected. Then you pull back along $f$ since you are interested in the domain. So you naturally consider the inverse images of $A$ and $B$. Then you just need to show they witness the disconnectedness of $X$. – Ittay Weiss May 29 '16 at 02:45
  • So we've already defined $A,B$ to be open sets that give us a separation of $f(X)$, and this makes sense. Why do we later write that $A = f(X) \cap U$, where $U \subset Y$ is open? – Irregular User May 29 '16 at 02:50
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    for no good reason. Try to ignore the proof you have in front of you, and instead work out the details yourself. All you want to show is that the inverse images of $A$ and $B$ form a separation of $X$. Can you give reasons for them being disjoint, non-empty, and why their union is $X$? The reasons are purely set-theoretic (the only place where you use topology here is in knowing the inverse images are open). – Ittay Weiss May 29 '16 at 02:55
  • Oh! So we just use the fact that inverse images behave nicely: $C \cup D = f^{-1}(A) \cup f^{-1}(B) = f^{-1}(A \cup B) = f^{-1}(f(X)) = X$. Similarly for $C \cap D = \varnothing$. And $C,D \neq \varnothing$ since $A,B \neq \varnothing$. Is this correct? – Irregular User May 29 '16 at 03:11
  • @IrregularUser: When we say "$A$, $B$ to be open sets that give us a separation of $f(X)$", we mean that $A$ and $B$ are open in $f(X)$ with respect to the subspace topology. –  May 29 '16 at 03:20
  • @Jack I get that $A$ cannot be the whole of $f(X)$ so we need to restrict it to some open set $U \subset Y$, but I can't see why introducing this new $U$ helps with the proof at all. EDIT: okay apparently I really don't understand that part since $U \subset Y$ instead of $U \subset f(X)$, so it's not really a restriction. – Irregular User May 29 '16 at 03:22
  • The reason we need $U$ is that we want to use the continuity of $f$ to show that $f^{-1}(A)$ is open in $X$. –  May 29 '16 at 03:30
  • If you don't understand where this $U$ is from, you should really take a look at the definition of subspace topology first. –  May 29 '16 at 03:33
  • $A$ is open in $f(X)$ (w.r.t. the subspace topology) means there is an open set $U$ in $Y$ such that $A=f(X)\cap U$. –  May 29 '16 at 03:34
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    @Jack Thanks for the reminder, think I got it now. So since $A, B$ are open in the subspace topology, there exists $U, V \in \mathcal{T}_Y$ such that $A = f(X) \cap U, B = f(X) \cap V$. Since $A \cup B = f(X) \cap (U \cup V)$, we have that $U \cup V = f(X)$. So, $C \cup D = f^{-1}(A) \cup f^{-1}(B) = f^{-1}(f(X) \cap U) \cup f^{-1}(f(X) \cap V) = (X \cap f^{-1}(U)) \cup (X \cap f^{-1}(U)) = X \cap (f^{-1}(U \cup V)) = X$ – Irregular User May 29 '16 at 03:42
  • @Jack Was that correct? – Irregular User May 29 '16 at 03:43