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## Homework Statement

As the title suggests. Rω is the space of all infinite sequences of real numbers.

The uniform topology is induced by the uniform metric, which is, on Rω, given with:

d(x, y) = sup{min{|xi - yi|, 1} : i is a positive integer}

## The Attempt at a Solution

I am trying to show that there is a subset of Rω, different from Rω and non empty, which is both open and closed to conclude that Rω is not connected in the uniform topology.

Let A be the set of all bounded infinite sequence of real numbers, and let x be an element of A. Then, for every integer i, xi <= M, where M is some real number.

To show that a is open in this topology, for any x from A there must exist some ε > 0 such that Bd(x, ε) is contained in A.

A point y is in Bd(x, ε) if sup{min{|xi - yi|, 1}} < ε. One must find an ε > 0 such that yi is a bounded sequence too, i.e. that the relation above implies that yi is bounded.

Now, let's take ε = 1. This implies that there is no i such that |xi - yi| >= 1, since if |xi - yi| would be equal or greater to 1 for some i, the supremum would equal 1. So, |xi - yi| < 1, for every i. If this is true, for any i yi must lie in the interval <xi - 1, xi + 1>. Since xi is bounded by M, it follows that yi is bounded by M + 1. Hence, A is open.

Now let's look at its complement, Rω\A, the set of all unbounded infinite sequences of real numbers. If we manage to show it's open, it follows that A is closed, i.e. clopen, and the problem is solved.

Let x be an unbounded sequence. One must find some ε > 0 such that Bd(x, ε) is contained in Rω\A. If ε = 1 again, and if xi is unbounded (so, for every real number a some xi such that xi > a), and if y is in Bd(x, ε), again for any i yi must lie in the interval <xi - 1, xi + 1>. Let a be a real number. Since there exists some xi such that xi > a, and since for xi + 2 there too exists some xj such that xj > xi + 2, and since for j yj must lie in <xj - 1, xj + 1>, we conclude that yj is greater than a. Hence, y is unbounded.

I'm really not sure about this last part, I hope at least I'll make someone laugh out loud if I wrote something stupid. :)

Thanks, and sorry it this is too long and tiring, but I tried to be precise.