Assignment 2–Fundamentals of Real Analysis – MATH 2027 7 Question 7 (Chapters 2 – 4). In this exercise, 2 marks are earned for each correct answer to the 10 state-
ments below (20 marks in total).
For each statement, decide if it is true or false. In order to obtain full marks for your answers, you need to give a short but irrefutable justification for those statements you believe to be true (you may either provide a short proof or quote results from lectures/textbook), or give a counterexample for those you consider false.
If (xn) ⊂ R is bounded, and (yn) is unbounded above, with yn ̸= 0 for all n, then (xn ) → 0. yn
(ii) (iii) (iv)
(v) (vi) F (vii) F(viii) F (ix)
If (xn) is monotone, then it must have a subsequence which is Cauchy. A closed set that contains every rational must be the whole R.
Every closed set must have a limit point.
A compact set C ⊂ R must have limit points.
Assume that C ⊂ R is a nonempty compact set and recall that I are the irrational numbers. Then C ∩ I cannot be compact.
If h, g are continuous in R, then the set
{x∈R : sin(h(x)2 +1)+5g(x)2 <−3, or cos(g(x)h(x))<0},
Every sequence (xn) ⊂ R has a monotone subsequence. If A ⊂ R is countable then A has no limit points.
If (xn) ⊂ R is a Cauchy sequence, then (yn) defined by yn := (−2)nxn for every n, is also a Cauchy sequence.
is closed. [10×2= 20 MARKS]
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True fix KEN
let M o InulEM Yu3km Fn no
ta no fn no
Inyyul arbitrary
A countable and
False take nu L
th and yn unbded hence
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monotone and Candy
false only
F 114 closed with
compact see Cvii
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False take C Then Cn I C
If compact
because ft
fzcx cog open because fz contin
s is finite intersection of open sets and hence open
deduce that s cannot be closed
since Stf IR