Real Analysis Exam 2

ℝ

Open: Every ε-neighborhood around any real number is contained in ℝ.
Closed: ℝ contains all of its limit points

ℚ

Not open: Let x=1. Every ε-neighborhood around 1 contains irrational numbers, so no neighborhood is completely contained in ℚ.
Not closed: √2 is a limit point of ℚ but √2 ∌ ℚ.

ℝ/ℚ

Not open: Let x = √2. Every ε-neighborhood contains rational numbers.

Not closed: 1 is a limit point of the irrational numbers, but 1 ∌ ℝ/ℚ.

𝕫

Not Open: Let x=1. Every ε-neighborhood around 1 contains non-integers.

Closed: The integers contain all their limit points (𝕫 has no limit points).

ℝ/𝕫

Open: For any non-integer x choose ε smaller than its distance to the nearest integer.

Not closed: 0 is a limit point of the non-integers but 0 ∌ ℝ/𝕫.

{ x ∋ ℝ| x > 0}

Open: For any x > 0 choose ε = x/2. Then (x-ε,x+ε) ⊆ (0,∞).

Not closed: 0 is a limit point but 0 is not in the set.

(0,1]

Not open: Let x=1. Every ε-neighborhood around 1 contains numbers greater than 1, which are not in the set.

Not closed: 0 is a limit point but 0 ∌ (0,1].

{1+1/4+1/9+…+1/n² | n ∋ ℕ}

Not open: Each point is isolated since the partial sums are increasing. Take n=1. Any neighborhood of 1 contains numbers below 1.

Not closed: π²/6 is a limit point because the partial sums converge to it but every partial sum is less than π²/6 so π²/6 ∌ the set.

lim 2x+4=8

Let ε > 0.

Let 𝛿 = ε/2.

If 0 < |x-2| < 𝛿, then |(2x+4)-8| = 2|x-2| < 2𝛿 = ε .

lim x³ =8

Let ε > 0.

|x³-8| = |x-2||x²+2x+4|

|x-2| < 1 so 1<x<3.

Then, x²+2x+4 < 9+6+4 = 19

Let 𝛿 = min {1, ε/19}.

If 0 < |x-2| < 𝛿, then |x³-8| < 19(ε/19) = ε.

lim |x| = 0

Let ε > 0 and 𝛿 = ε.

If 0 < |x-0| < 𝛿 then ||x| - 0| = |x| < ε.

g(x) = ∛x at c=0

Let ε > 0 and 𝛿 = ε³.

If |x-0| < 𝛿 then |∛x| = |x|^1/3 < (ε³)^1/3 = ε.

So g is continuous at 0.

g(x) = ∛x at c=1

Let a = ∛x, x = a³.

x-1 = a³-1 = (a-1)(a²+a+1)

Since a² + a +1 = (a+1/2)²+3/4 ≥ 3/4,

|∛x-1| = |x-1| / (a²+a+1) ≤ (4/3) |x-1|.

Let ε > 0 and 𝛿 = 3ε/4.

If |x-1| < 𝛿 then |g(x)-g(1)| ≤ (4/3)|x-1| < ε.

Therefore, g is continuous at 1.