Contrapositive of theorem and issue with proof

[psie]

TL;DR Summary

I'm stuck on a contrapositive of a theorem and a logical issue with a proof. It concerns the statement that the image of a connected set is connected under a continuous map.

Consider the following theorem:

Theorem Suppose that $f:X\to Y$ is a continuous map between two topological spaces $X$ and $Y$. Then $f(X)$ is connected if $X$ is.

First, I don't know how to take the contrapositive of this statement. I'm not sure if the opening hypothesis, that is, $f$ is continuous, remains outside. Because the way this is proved in the text I'm reading is by assuming $f(X)$ is disconnected and then using continuity to show that $X$ is also disconnected. Since the author uses the continuity assumption, it seems like that is not a part of the statement that one takes the contrapositive of.

Yet, I saw someone claim,

Claim If a function has a connected domain $X$ and a disconnected range $f(X)$, then the function is not continuous on $X$.

So in this claim it seems like that the continuity is actually negated as well. I'm confused and I'm now doubting the validity of the proof of the theorem. What is correct and what isn't?

 

[.Scott]

A: A function f(X) is a continuous map between two topological spaces X and f(X)
B: f(X) is connected
C: X is connected

If C and Not B, then Not A.

So the original sentence is When A: Then B, if C.
This can be rewritten: If A, then (if C then B)
The contrapositive: If not (if C then B) then not A.

(if C then B) is false only when C is true and B is false
so: not (if C then B) == ( C and not B)
Thus: If not (if C then B) then not A == if (C and not B) then not A

QED

 

[PeroK]

So in this claim it seems like that the continuity is actually negated as well. I'm confused and I'm now doubting the validity of the proof of the theorem. What is correct and what isn't?

The theorem itself is of the form:

Suppose $f$ is continuous, then property A holds.

The contrapositive of that is:

If property A does not hold, then $f$ is not continuous.

That's probably the answer for the contrapositive of the theorem.

In this case, property A is also an "if-then":

If $X$ is connected, then $f(X)$ is connected.

The contrapositive of property A is:

If $f(X)$ is not connected, then $X$ is not connected. But, that's not the contrapositive of the theorem.

 

[psie]

Hmm, ok. I'm still uncertain about that proof. Is it correct to use the continuity of the function when proving the contrapositive of the theorem?

 

[PeroK]

Hmm, ok. I'm still uncertain about that proof. Is it correct to use the continuity of the function when proving the contrapositive of the theorem?

What I said was that there is a difference between the contrapositive (statement) of the theorem and the contrapositive (statement) of a condition or property used in the theorem.

In that sense, we are arguing about a definition of what precisely the contrapositive means in this case. There's no error in either argument that is not based on an interpretation of the question.

 

[Evgeny.Makarov]

Let $A$, $B$ and $C$ be propositions. Then the following formulas are equivalent.
\begin{align*}
&A\to (B\to C)\\
&A\land B\to C\\
&B\to (A\to C)\end{align*}Therefore there are many ways to construct equivalent statements using contrapositive. Below $F\equiv G$ means that $F$ and $G$ are equivalent.
\begin{align*}
A\to (B\to C)&\quad\equiv\quad\neg(B\to C)\to\neg A&&\equiv\quad B\land\neg C\to\neg A\tag{1}\\
A\to (B\to C)&\quad\equiv\quad A\to(\neg C\to\neg B)\tag{2}\\
A\land B\to C&\quad\equiv\quad\neg C\to\neg(A\land B)&&\equiv\quad\neg C\to\neg A\lor\neg B\tag{3}\\
B\to (A\to C)&\quad\equiv\quad\neg(A\to C)\to\neg B&&\equiv\quad A\land\neg C\to\neg B\tag{4}\\
B\to (A\to C)&\quad\equiv\quad B\to(\neg C\to\neg A)\tag{5}\end{align*}Usually contrapositive is applied when the assumption and the conclusion of the implication are atomic rather than compound propositions. This means that proving formulas at the end of lines (1) and (4) is somewhat unusual but still valid way to prove $A\to(B\to C)$. I believe a more common way is to leave one of the assumptions $A$ or $B$ unchanged and then take the contrapositive of the remaining implication as in lines (2) and (5).

In this situation if we denote "$f$ is continuous" by $A$, "$X$ is connected" by $B$ and "$f(X)$ is connected" by $C$, then the text proved formula in (2). The second claim in post 1 is (1).