Countability

[匿名]

2015-03-07 2:27 am

Let T be a nonempty subset of the interval (0, 1).
If every finite subset {x1, x2, . . . ,xn} of T (with no two of x1, x2, . . . ,xn equal) has the property that (x1)^2+(x2)^2+(x3)^3+...+(xn)^2<1,
then prove that T is a countable set.

Thanks for solving.

回答 (1)

Andrew

2015-03-07 9:54 pm

✔ 最佳答案

I had some fun thinking about this one - the key idea is that there cannot be too many large numbers - here is a way to prove it.

First, note that there cannot be two numbers larger than sqrt(1/2) - or otherwise the constraint is broken.

Second, note that there cannot be three numbers larger than sqrt(1/3) - or otherwise the constraint is broken.

... now we see the obvious pattern.

Write the set T as a union of countable set of finite sets T_n (n > 1) such that all elements in T_n is less than sqrt(1/n) - by the argument above, each of the set is finite.

T as a countable union of finite set is countable!

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