Real analysis, math homework help

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Mathematics

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metric spaces

Just exercise 1,2,5,8

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Exercise 1. Let (M, p) be a metric space and define the metric p on R by p(x, y) = |x - y Assume Assume f,g:M + R are chitinous functions. a) Show that cf is continuous for all constants c. b) Show that f + g is continuous. c) Show that fg is continuous. = Exercise 2. a) suppose (M,p) is a metric space, and define p(x,y) P1(x,y) 1+ p1(x, y) b) Show that infinitely many metrics can be defined on any set M with more than one member. Exercise 3. Prove that if p and o are both metrics for a set M, the pto is also a metric for M. Exercise 4. Let p(x,y) = Prove that p is a metric on 0,00). Exercise 5. Let (M, p) be a metric space. Prove that |p(x,y) - P(x,z)
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Explanation & Answer

The solutions are ready. Please read them and ask if something is unclear.It is probably a typo in q.2a.The docx and pdf files are the same.

1. A function 𝑓: 𝑀 → 𝑀1 is called continuous in a point 𝑚 ∈ 𝑀 if it is defined in some
neighborhood of 𝑚 and
∀𝜀 > 0 ∃𝛿(𝜀): ∀𝑥 ∈ 𝐵𝛿(𝜀) (𝑚) ⇒ 𝑓(𝑥) ∈ 𝐵𝜀 (𝑓(𝑚)).
Here 𝐵𝑟 (𝑦) denotes the open ball centered at 𝑦 with the radius 𝑟 in the corresponding
metric space, 𝐵𝑟 (𝑦) = {𝑥 ∈ 𝑀: 𝜌(𝑥, 𝑦) < 𝑟}.

a. It is evident for 𝑐 = 0. For nonzero 𝑐 estimate |𝑐𝑓(𝑥) − 𝑐𝑓(𝑚)| = |𝑐||𝑓(𝑥) − 𝑓(𝑚)|,
𝜀

which is < 𝜀 for ∀𝑥 ∈ 𝐵𝛿(𝜀) (𝑚) (because for those 𝑥 |𝑓(𝑥) − 𝑓(𝑚)| < 𝑐).
𝑐

b. Given 𝜀 > 0 and having 𝛿𝑓 (𝜀) and 𝛿𝑔 (𝜀) from the definition of continuity, use
𝜀
𝜀
𝛿(𝜀) = min (𝛿𝑓 ( ), 𝛿𝑔 ( )) > 0
2
2
which is sufficient:
|(𝑓(𝑥) + 𝑔(𝑥)) − (𝑓(𝑚) + 𝑔(𝑚))| ≤ |𝑓(𝑥) − 𝑓(𝑚)| + |𝑔(𝑥) − 𝑔(𝑚)| <

𝜀 𝜀
+ =𝜀
2 2

for all 𝑥 ∈ 𝐵𝛿 (𝑚).

c. The required estimate is
|(𝑓 ∙ 𝑔)(𝑥) − (𝑓 ∙ 𝑔)(𝑚)| = |𝑓(𝑥)𝑔(𝑥) − 𝑓(𝑥)𝑔(𝑚) + 𝑓(𝑥)𝑔(𝑚) − 𝑓(𝑚)𝑔(𝑚)| =
= |𝑓(𝑥)(𝑔(𝑥) − 𝑔(𝑚)) + (𝑓(𝑥) − 𝑓(𝑚))𝑔(𝑚)| ≤
≤ |𝑓(𝑥)||𝑔(𝑥) − 𝑔(𝑚)| + |𝑔(𝑚)||𝑓(𝑥) − 𝑓(𝑚)|.

The differences |𝑓(𝑥) − 𝑓(𝑚)|, |𝑔(𝑥) − 𝑔(𝑚)| are arbitrary small for 𝑥 close to 𝑚.
The factor |𝑓(𝑥)| is bounded in some neighborhood of 𝑚:
|𝑓(𝑥)| ≤ |𝑓(𝑥) − 𝑓(𝑚)| + |𝑓(𝑚)|,
∀𝑥 ∈ 𝐵𝛿(1) (𝑚) ⇒ |𝑓(𝑥)| ∈ 𝑈1 (𝑓(𝑚)) ⇒ |𝑓(𝑥)| ≤ 1 + |𝑓(𝑚)|.
1

More formally, for a given 1 > 𝜀 > 0 denote 𝑒 = 2(1+|𝑓(𝑚)|+|𝑔(𝑚)|) and pick
𝛿(𝜀) = min (𝛿𝑓 (𝑒), 𝛿𝑔 (𝑒)) > 0. Therefore
|(𝑓 ∙ 𝑔)(𝑥) − (𝑓 ∙ 𝑔)(𝑚)| ≤ |𝑓(𝑥)||𝑔(𝑥) − 𝑔(𝑚)| + |𝑔(𝑚)||𝑓(𝑥) − 𝑓(𝑚)| ≤
≤ (1 + |𝑓(𝑚)|)𝑒 + |𝑔(�...


Anonymous
Really useful study material!

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