Question

proof of sandwich theorem ?​

Answer 1

Answer:

Hope it helps

Step-by-step explanation:

Suppose we have the following statement (an)→ℓ,(bn)→ℓ and we have

an≤cn≤bn

then (cn)→ℓ.

I think I have a proof which goes as follow ;

an≤cn≤bn≤ ⇒0≤cn−an≤bn−an

, as the terms are all larger than 0, taking the absolute value will not change any of the signs of the inequalities. So we have

0≤|cn−an|≤|bn−an|.

Now consider

|bn−an|=|(bn−ℓ)+(ℓ−an)|≤|bn−ℓ|+|an−ℓ| (by triangle inequality).

Using the definition of a sequence tending to a value, if (an)→ℓ then ∃N1∈N s.t ∀n>N,|an−ℓ|<ϵ ,∀ϵ>0. We do the same for (bn) but replacing N1 with N2 and using the same ϵ without loss of generality. So we can now say that

|bn−an|≤|bn−ℓ|+|an−ℓ|<2ϵ.

So we have

0≤|cn−an|≤|bn−an|<ϵ0, where ϵ0=2ϵ.

So we can conclude (using sandwich theorem for null sequences) that (cn−an)→0⇒(cn)→(an)⇒(cn)→ℓ since (an)→ℓ.

Answer 2

Answer:

Borsuk-Ulam theorem

proof of ham sandwich theorem. This proof uses the Borsuk-Ulam theorem, which states that any continuous function from Sn to ℝn maps some pair of antipodal points to the same point.