1. Prove that if {x,21 converges to x, then { x,„k} converges to xk for all natural numbers k.ritr Use mathematical induction and the previous theorems we have proven.) 2. Let a, b E R. Prove the...

1. We are given that xn → x. We have to prove xkn → xk. Using Principle of Mathematical Induction let’s first
see that this statement is true for k = 2 and then proceed. So to prove that the statement is true for k = 2 we
have to show that x2n → x2. This means that given � > 0, there is an n ∈ N, such that for all n > N we have
|x2n − x2| < �. But
|x2n − x2| = |xn − x| · |xn + x|
= |xn + x| · �
[
∵ xn → x
]
≤ M · �
Since {xn} is a convergent sequence therefore its bounded. So I can take |xn + x| ≤ M . Another way of
proving this easily is the following:
Note that f(x) = x is a continuous function. And [f(x)]n is a continuous function for every
n ∈ N. In particular [f(x)]2 = x2 is a continuous function. And note that if f is continuous and
{xn} → x then f(xn) → f(x).
So we have shown that for k = 2, the result is true. Now assume that for all numbers ≤ k ∈ N the result is
true. That is xkn → xk. We have to show that xk+1n → xk+1. Now consider
|xk+1n − xk+1| = |xkn − xk| · |xn + x|+
(
xnx
k − xknx
)
≤ � ·M + xnx · |xk−1 − xk−1n |
≤ M · �+K · �
≤ υ
Hence xk+1n → xk+1.
2. (a) Since a, b ∈ R, we have
• (i) a = b or
• (ii) a > b or
• (iii) a < b.
Now clearly a = b, is not possible. Because if a = b, then our hypothesis a ≤ a+ � will not hold for any � > 0.
Moreover, a > b is also not possible, since this doesn’t satisfy a ≤ b+ �. So the only choice left is a ≤ b.
2. (b) Note that 0 ≤ a− b ≤ � for every � > 0 implies that b ≤ a ≤ b+ �. Now by considering the first inequality we
get b ≤ a and from a ≤ b+ � from (a) implies a ≤ b. Hence a = b.
3. (a) Let ` = inf B....