Question about open sets?

Question

Let A be the set of all sequences {s_n} (n=1 to infinity) in L^2 (L^2, being the typical definition, = {{s_n} (n=1 to infinity) : the sum n=1 to infinity of s_n^2 is finite} such that the sum of s_n^2 is less than 1.  Prove that A is an open subset of L^2.

I figure that I am using this definition to prove this....

Let M be a metric space.  We say the subset A of M is open of M if for all x in A there exists r>0 such that the entire open ball B(x;r) is contained in A.

B(x,r) = {x in A: rho(x,a)<r}

Since we are told we are in L^2, I'm trying to figure out how to use that....

Suleiman · Accepted Answer

[Note: you've defined B(x,r) as {x ∈ A | ρ(x,a) < r};
it should be {a ∈ A | ρ(x,a) < r}].

Well the elements of ℓ² with length less than 1 are the
elements whose distance from 0 is less than 1, so
A = B(0,1).

Now, to see that open balls are indeed open, consider
x ∈ B(x₀,r): we need to find an open ball centered
at x contained in B(x₀,r).
We have ρ(x,x₀) < r, and hence 0 < r - ρ(x,x₀).
Take B(x, r - ρ(x,x₀)); for y ∈ B(x, r - ρ(x,x₀)) we have
by the triangle inequality that
ρ(y,x₀) ≤ ρ(x,y) + ρ(x,x₀)
< (r - ρ(x,x₀)) + ρ(x,x₀)
= r,
and so B(x, r - ρ(x,x₀)) ⊂ B(x₀,r).

→ Open balls are open sets, and therefore A = B(0,1) is open.

Trending News

Question about open sets?

1 Answer