Metric Spaces

Question 1

v = (v1,…, vn),w = (w1,…,wn) are vectors in Rⁿ
then we set
=

Answer 1

= ᵢ₌₁Σⁿvᵢwᵢ

Question 2

Define a length of a vector

Answer 2

||v|| = ¹/²

Question 3

What does it mean for a sequence of vectors in Rⁿ to converge to a vector w ϵ Rⁿ?

Answer 3

If (vᵏ)ₖϵₙ is a sequence of vectors in Rn (so vᵏ = (vᵏ₁ ,…, vᵏₙ)) we say (vᵏ)ₖϵₙ converges to
w ϵ Rⁿ if for any ε > 0 there is an N > 0 such that for all k ≥ N we have ||vᵏ - w|| < ε.

Question 4

If f: Rⁿ → R and a ϵ Rⁿ then we say that f is continuous at a if ….

Answer 4

If for any ε > 0 there is a 𝛿 > 0 such that |f(a) - f(x)| < ε whenever ||x - a|| < 𝛿

Question 5

Suppose that (vᵏ)ₖ≥₁ is a sequence in Rⁿ which converges to w ϵ Rⁿ and also to u ϵ Rⁿ. Then w =

Answer 5

w = u. That is, limits are unique

Question 6

Suppose that (vᵏ)ₖ≥₁ is a sequence in Rⁿ which converges to w ϵ Rⁿ and also to u ϵ Rⁿ. Then w = u. That is, limits are unique

Prove it

Answer 6

We prove this by contradiction: suppose w ≠ u
Then d = ||w - u|| > 0, so since (vᵏ) converges to w we can find an N₁ ϵ N s.t for all k ≥ N₁ we have ||w - vᵏ|| < d/2. Similarly since (vᵏ) converges to u we can find an N₂ ϵ N s.t for all k ≥ N₂ we have ||vᵏ - u|| < d/2.
Bu then if k ≥ max{N₁, N₂} we have
d = ||w - u|| = ||(w - vᵏ) + (vᵏ - u)|| ≤ ||w - vᵏ|| + ||vᵏ - u|| < d/2 + d/2 = d
contradiction

Question 7

Let X be a set and suppose that d : X x X → R
Define a distance function
(X cross X)

Answer 7

Then we say that d is a distance function
on X if it has the following properties: For all x, y, z ϵ X
1) Positivity: d(x,y) ≥ 0 and d(x,y) = 0 iff x = y
2) Symmetry: d(x,y) = d(y,x)
3) Triangle Inequality: If x,y,z X then we have
d(x,z) ≤ d(x,y) + d(y, z)

Question 8

What is a metric space?

Answer 8

We will write a metric space as a pair (X,d) of a set and a distance function d : X x X → R≥₀ satisfying the distance axioms

Question 9

If (X,dₓ) is a metric space and A ⊆ X then we set

diam(A) =

Answer 9

diam(A) = sup{d(a1,a2) : a1,a2 ϵ X} ϵ R≥₀ ∪ {∞}