I'm doing the following exercise from Just/Weese:
Some thoughts: To show that the statement is not provable from $ZF$ I could either show that it implies the axiom of choice or Tychonoff or I could assume that it is provable from $ZF$ and produce a contradiction. Since I'm give a statement that is known not to imply $AC$ and I'm told to use it, I will do so. The idea is that if the second statement in the exercise is provable from $ZF$ then the first statement implies Tychonoff's theorem which is the desired contradiction.
Proof: Let (*)"For every indexed family of non-empty sets there is a family of compact Hausdorff topologies." be provable from $ZF$. Assume that (**)"The product of compact Hausdorff families is compact.". Let $X_i$ be an arbitrary collection of
sets compact spaces . Then by (*) we can turn them into a compact Hausdorff family and by (**) the product is compact. Hence Tychonoff's theorem holds which is equivalent to $AC$. But then (**) would be equivalent to $AC$.$\Box$
Some more thoughts: Given a single arbitrary set it is always possible to endow it with a compact Hausdorff topology: First put the discrete topology on it and then use one-point compactification to make it into a compact space.
The reason why this cannot be applied to an infinite family of sets is that for each set one needs to choose a bijection between the set and the set union one point. And to do so one needs $AC$.
Can you tell me if I have it right? Thank you.