I am having doubts about Theorem 2.36 pasted below. I was able to follow all the steps individually, but I don’t see how this is a proof by contradiction. It seems to be it is a proof by contrapositive.
That is because, we are assuming intersection of all $\{ K_{\alpha} \}$ is empty, then showing that intersection of an arbitrarty finite subcollection $K_1 \cap K_{\alpha_{1}} \cap \cdot\cdot\cdot \cap K_{\alpha_{n}} = \phi$.
What am I missing here? Why is this not proof by contrapositive?
On
To make things clearer, first, proof by contrapositive is a special case of proof by contradiction. That is, from a logical point of view, if you want to prove $\lnot q \implies \lnot p$ is logically equivalent to $p \implies q$, then you will have to use proof by contradiction with both directions.
Now about the theorem, if you try to formalize it, you will end up with something like this:
Let $S$ be a subset of $\mathcal{P}(X)$ indexed by $\Lambda$, i.e. $S = \{ K_{\alpha} : \alpha \in \Lambda \}$ where $\Lambda$ is an indexing set.
We want to prove the statement $p \wedge r \implies q$, where $p$ is the statement "$K_\alpha$ is compact for all $\alpha \in \Lambda$", $r$ is the statement "the intersection of every finite subcollection of $S$ is nonempty", and $q$ is the statement "$\bigcap\limits_{\alpha \in \Lambda} K_{\alpha} \neq \phi$".
In the proof, we assumed $\lnot q$ and $p\wedge r$, then we concluded that $\lnot p$. Since by conjunction elimination $p \wedge r \implies p$, we conclude $p \wedge \lnot p$ which is a contradiction, then we conclude $q$ by reductio ad absurdum.
Looking at it from another perspective, by taking the contrapositive of $p \wedge r \implies q$ which is $\lnot q \implies \lnot p \vee \lnot r$, we start by assuming $\lnot q$, then we assume $r$ as in the proof of the theorem to conclude $\lnot p$. Thus, we have $\lnot q \wedge r \implies \lnot p$. However, $a \implies b$ is logically equivalent to $\lnot a \vee b$. Thus, $\lnot q \wedge r \implies \lnot p$ is logically equivalent to $\lnot(\lnot q \wedge r) \vee \lnot p$ which is equivalent to $q \vee \lnot r \vee \lnot p$, i.e. $q \vee (\lnot r \vee \lnot p)$ by the associativeness of disjunction. Hence, by turning this back into implication, we obtain $\lnot q \implies \lnot p \vee \lnot r$.
So you can look at it in both ways. It is a proof by contradiction and a proof by contrapositive too.
Whether one sees it as a proof by contradiction or as a proof of the contrapositive depends on whether one assumes that the $K_\alpha$ have the finite intersection property. If one makes that assumption, then the argument is indeed a proof by contradiction: the further assumption that the intersection of all of the $K_\alpha$ is empty leads to a conclusion that contradicts that contradicts the starting assumption. That further assumption must therefore be false.
If one does not make that initial assumption, the same argument proves that the $K_\alpha$ do not have the finite intersection property and thus demonstrates the contrapositive of the stated result. I prefer to view it as a proof of the contrapositive, but both views are logically defensible, and Rudin evidently chose to adopt the other view.
Added: The difference is in the logical structure of the proof of an implication $A\implies B$. In a direct proof we assume $A$ and somehow derive $B$. In a proof of the contrapositive we assume $\neg B$ and somehow prove $\neg A$; this direct proof of the contrapositive of course establishes the logically equivalent original implication $A\implies B$.
In a proof by contradiction we assume $A$ and $\neg B$ and somehow derive a contradiction. This shows that $A$ and $\neg B$ cannot both be true, so if $A$ is true, then $\neg B$ must be false and therefore $B$ must be true. It sometimes happens in the in deriving the contradiction we never really use the assumption that $A$ is true: we actually use only the assumption $\neg B$ and get our contradiction (with the assumption $A$) by deriving $\neg A$ from $\neg B$. In that case, which is what we have here, we’ve presented what could have been a direct proof of the contrapositive in the logical form of a proof by contradiction, and technically that form makes it a proof by contradiction. It just didn’t have to be one. We didn’t use the assumption $A$ at any point in the actual mechanics of the argument: it served only to be contradicted.