Question

In: Advanced Math

Let q and p be natural numbers, and show that the metric on Rq+p is equivalent...

Let q and p be natural numbers, and show that the metric on R^q+p is equivalent to the metric it has if we identify R^q+p with R^q × R^p.

Expert Solution

We want to show, the metric topology on is equivalant to product topology of . We can identify points of and in natural way. Now pick . Take an open ball arround of radius . Now it contains points , such that . That is same as saying, Now, and . Take balls and of radius arround and . And consider . Now, belong to the product topology of .

contains all points such that and . Then in , . So, .

Next we need to show for , is a ball centered at of radius , and is a ball centered at of radius , we can find a real number , such that . Now, for , and for ,. With out loss of generality, let be minimum of . Then take . Then . That imply and . So, for this , .

So these two topologies are equivalant.

Colby Messinger answered 1 year ago

Let p and q be propositions. (i) Show (p →q) ≡ (p ∧ ¬q) →F (ii.)...

Let p and q be propositions. (i) Show (p →q) ≡ (p ∧ ¬q) →F (ii.) Why does this equivalency allow us to use the proof by contradiction technique?

Let p and q be any two distinct prime numbers and define the relation a R...

Let p and q be any two distinct prime numbers and define the relation a R b on integers a,b by: a R b iff b-a is divisible by both p and q. For this relation R: Prove that R is an equivalence relation. you may use the following lemma: If p is prime and p|mn, then p|m or p|n. Indicate in your proof the step(s) for which you invoke this lemma.

1. Show that the argument (a) p → q q → p therefore p...

1. Show that the argument (a) p → q q → p therefore p V q is invalid using the truth table. ( 6 marks ) (b) p → q P therefore p is invalid using the truth table. ( 6 marks ) (c) p → q q → r therefore p → r is invalid using the truth table. ( 8 marks )

Show that if P;Q are projections such that R(P) = R(Q) and N(P) = N(Q), then...

Show that if P;Q are projections such that R(P) = R(Q) and N(P) = N(Q), then P = Q.

Two assets P and Q, E(rp) = 10%, σp = 10%, E(rq) = 15%, σq =...

Two assets P and Q, E(rp) = 10%, σp = 10%, E(rq) = 15%, σq = 25%. The risk-free rate of interest (on bills) is 5%. Now, let’s allow borrowing on margin. Investors can buy P and Q on margin (but not both). The borrowing rate is the risk-free rate. State whether the following statements are true or false and give reasons: (a) “Investor who prefers P to Q must be risk-averse.” (b) “Investor who prefers Q to P cannot...

If R is a P.I.D. then an R-module Q is injective if and only if rQ=Q...

If R is a P.I.D. then an R-module Q is injective if and only if rQ=Q for every nonzero r in R.

(a) Let <X, d> be a metric space and E ⊆ X. Show that E is...

(a) Let <X, d> be a metric space and E ⊆ X. Show that E is connected iff for all p, q ∈ E, there is a connected A ⊆ E with p, q ∈ E. b) Prove that every line segment between two points in R^k is connected, that is Ep,q = {tp + (1 − t)q | t ∈ [0, 1]} for any p not equal to q in R^k. C). Prove that every convex subset of R^k...

Let T∈ L(V), and let p ∈ P(F) be a polynomial. Show that if p(λ) is...

Let T∈ L(V), and let p ∈ P(F) be a polynomial. Show that if p(λ) is an eigenvalue of p(T), then λ is an eigenvalue of T. Under the additional assumption that V is a complex vector space, and conclude that {μ | λ an eigenvalue of p(T)} = {p(λ) | λan eigenvalue of T}.

(Connected Spaces) (a) Let <X, d> be a metric space and E ⊆ X. Show that...

(Connected Spaces) (a) Let <X, d> be a metric space and E ⊆ X. Show that E is connected iff for all p, q ∈ E, there is a connected A ⊆ E with p, q ∈ E. b) Prove that every line segment between two points in R^k is connected, that is Ep,q = {tp + (1 − t)q | t ∈ [0, 1]} for any p not equal to q in R^k. C). Prove that every convex subset...

For p, q ∈ S^1, the unit circle in the plane, let d_a(p, q) = min{|angle(p)...

For p, q ∈ S^1, the unit circle in the plane, let d_a(p, q) = min{|angle(p) − angle(q)| , 2π − |angle(p) − angle(q)|} where angle(z) ∈ [0, 2π) refers to the angle that z makes with the positive x-axis. Use your geometric talent to prove that d_a is a metric on S^1.