Problem 12\_June\_2015\_1

Let and be functions from such that whenever . Show that is continuous.

Problem 12\_June\_2015\_2

Let be monotonically increasing. Suppose , and . Show that there exists such that

Problem 12\_June\_2015\_3

Let power set of , for . [ stands for the product of first positive integers]. Let be determined by the condition . If and are independent random variables on show that at least one of them is almost surely constant.

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Solution for 2nd problem : Consider the non-empty set . We claim that is such that Suppose Clearly Let be such that Since , which contradicts the increasing property since If , by the supremum property there exist such that which is again a contradiction to the increasing property.Hence and clearly

p.s There has been some writing problem in my earlier comment so i am reposting it. Also it would be good if we have a preview option(by installing plugin) so that we can see our comment before posting.

Prahlad,

You are on the right track and your second part is correct. It shows that f(m) cannot be less than m. Why can’t it be grater than m?

Congratulations! You have solved it. – Proposer

can second problem be solved if f is not continious

Yes!

Consider the non-epmty set f(x)>x,x\in[a,b]\}$. We claim that is such that Suppose Clearly be such that

. If such that and , which is again a contradiction since . Hence and clearly