How can I think of counterexamples of abstract mathematical objects?

Question

I noticed that I have an intuition for theory (ie making connections between theorems, lemmas, and proving related statements).

However, when a problem gives me the choice between proving or finding counter examples, I immediately get lost. I have no idea how to begin creating concrete mathematical objects that would lead to a counter example.

Does anyone have any advice from past experience from the beginning of their mathematical career on how to develop comfort with coming up with examples and counter examples? It’s really frustrating and I feel very stupid.

Answer 1

I tend to think of proving theorems and finding counterexamples as two sides of the same coin. I try to prove it, and if I find that I can't, I figure out what's stopping me. I then find an object which exhibits that property. E.g. if trying to prove 'all subgroups of $\mathbb{Z}$ are isomorphic to $\mathbb{Z}$' the thing that trips you up is you end up having to assume the subgroup has a minimum positive element. So to find a counterexample, you look for a subgroup without a minimum positive element.

Answer 2

The best advice, I think is just to continue working on problems.

If there's a counter-example to some claim; then obviously it breaks down somewhere. It might help to break up the claim into components and see what might be suspsect. This "suspiciondar" just comes from experience with whatevever material you are working with. Just pick apart the theorem/claim and try reconstructing it.

Maybe an example will help.

Claim (false): Let A be a nonempty Set of Reals. Then,

sup(A) ≠ inf(A)

where sup and inf refer to the least upper bound and greatest lower bound respectively.

How do we go about finding a counter-example? Why wouldn't this be true?

One way is to consider when is it true. Well, if the cardinality of A is > 1. |A| > 1, it must be true as the Reals are connected. To see this Take two distinct elements in A, a and b. Without loss if generality If a > b then sup(A) ≥ a and inf(A) ≤ b by definition. So sup(A) ≠ inf(A). Ok, so the claim is true if A has more than one element.

So, let's consider when |A| = 1. A = {a} for some real number A.

Then inf(A) = a = sup(A).

A counterexample to our claim would be:

A = {1}