Finding a counter example

Many times we encounter problems in Mathematics which is a complete conjecture, about which we have no idea that whether it’s true or false. For example: Is it true that X happens in Y? etc ; This’s how Most problems in Pure Mathematics arises in real life, and they are quite beautiful in a strange, pleasant, and brutal sort of ways. Many a times I have been asked by different people at different points of time that how to approach such problem. Well, the best way to approach such problem is to assume first that the conjecture is true and let’s go on proving this, because proving a true result is often much easier than finding a counter example for the false one, but for average standard problems, even the trouble for finding a counter example can be abated largely if we are careful. The basic idea is: Assume that the conjecture is true and go on proving this; very carefully, taking care of all the steps and logic involved; if it is provable, i.e. if it’s true, you will definitely find some direction, at least you wont encounter a situation where Mathematics will surrender, and you will find it obvious to take a further step that is mathematically consistent and logical – this you’ll definitely find if the statement is not provable (or false). It’s better to present this idea with an example. Let me talk about a good problem that people encounter in the first course of Topology: Is [0,1] compact in co-countable topology? When I faced this problem, I asked so many questions related to this (as I always ask) even before I begin to prove this, but the most important one was: Is [0,1] compact in co-finite topology? Yes; it is; indeed if there’s an open cover for [0,1] and we pick just one member, it will leave at most finitely many points outside; now we will choose only those cover that contains those points (do make sure that such open sets exists); needless to say that at most finitely many such covers will be needed. I looked for the same argument in co-countable topology and got in to trouble. Well, where was the problem? If we choose a member from the open cover this time, it wont leave only finitely many points this time – here’s the key. Whereupon I looked for a family of open sets indexed by Natural numbers (or any countable set – you know, it doesn’t matter) such that the complement of every successor is contained in the predecessor. The first thing in front of me was the familiar sequence {1/n} So, I constructed the family with members U(k) = R \ {1/n ; n >= k} i.e. U(1) will be the set R \ {1, 1/2, 1/3, …}, U(2) = R \ {1/2, 1/3, …} Note that this family is an open cover for [0,1] (why?) Now, this is easy to see that no finite sub-collection can cover [0,1] for if p (say) is the largest index of the sub-collection then 1/(p+1) wont belong to the union. Soon, I found out another proof as well which borrows the idea from the above proof but is very clever: Just look at the sequences I removed from each of the members, they are closed set, indeed, and are nested, with empty overall intersection, that means if [0,1] had been compact it would have violated the famous finite intersection property.
Although the above strategy is very useful, but it also depends upon our approach; sometimes we proceed in a way from where we can’t get a straight forward hint like above. To sum up, I wont deny Yuri Manin that proof is much easier but proving is difficult – it involves experience, speculation, luck, and so many other things.