Although I have proof with me that Tynhonoff theorem implies AC. But I have some difficulties with it:
1. Do we define topology on empty set. If not then in proof of Tynhonoff theorem implies AC we don't need to prove anything as below:
Let (Xa) be collection of sets. Now we endow each Xa with indiscrete topology. Each Xa compact so there product compact and being topological space nonempty and we are done.
Am I right in saying like this?
2. If we can define topology on empty set, then I am not able to see why Tychonoff theorem should imply AC. May be all countable and further product of spaces are empty and have topology defined on empty set. Am topology on empty set is always compact as power set of empty set is finite.
Please help me in understanding what actually is going on here.
The proof is not saying "Oh, since the product of the space is a topological space, it is non-empty by definition".
The proof, instead, begins with a family of pairwise disjoint non-empty sets $A_i$, then takes the product of $X_i=A_i\cup\{i\}$. The product is never empty since we have a choice function $f(X_i)=i$. Moreover, since we may assume the index set is infinite, and each $A_i$ has more than one element, there are infinitely many elements in that product.
The compactness of the product ensures that there is a choice function from the $A_i$'s.
See more on Wikipedia.