Exercise 6.4.1

August 19, 2020 Issa Rice Leave a comment

Exercise statement

Prove Proposition 6.4.5.

Proposition 6.4.5 (Limits are limit points). Let $(a_n)_{n=m}^\infty$ be a sequence which converges to a real number $c$ . Then $c$ is a limit point of $(a_n)_{n=m}^\infty$ , and in fact it is the only limit point of $(a_n)_{n=m}^\infty$ .

Hints

None.

How to think about the exercise

This is a straightforward exercise so I don’t have anything to say.

Model solution 1

Let $(a_n)_{n=m}^\infty$ be a sequence which converges to a real number $c$ . To show that $c$ is a limit point of this sequence, let $\varepsilon > 0$ be a real number and let $N' \geq m$ be an integer. Since the sequence converges to $c$ , we know that there is some $N \geq m$ such that $|a_n - c| \leq \varepsilon$ for every $n \geq N$ . We want to find some $n \geq N'$ such that $|a_n - c| \leq \varepsilon$ . We can take $n := N+N'$ . Then we see that $n \geq N'$ , and since $n \geq N$ we have $|a_n-c|\leq \varepsilon$ as required.

Next we show that $c$ is the only limit point of the sequence. Suppose for sake of contradiction that there there is another limit point $c' \ne c$ . Put $\varepsilon := |c-c'|/3$ . Since $c\ne c'$ , we know that $\varepsilon$ is positive. Since the sequence converges to $c$ , there is some $N \geq m$ such that $|a_n-c|\leq \varepsilon$ for all $n\geq N$ . So fix this $N$ . Since $c'$ is a limit point, there exists an $n \geq N$ such that $|a_n - c'| \leq \varepsilon$ . Fix this $n$ . Since $n \geq N$ , we also know that $|a_n-c|\leq \varepsilon$ . But now we have

$|c-c'| = |c-a_n+a_n-c'| \leq |c-a_n| + |a_n-c'| \leq 2|c-c'|/3$

which is a contradiction since $|c-c'| > 0$ .

Model solution 2

Here’s an alternative way to prove the second part of this exercise.

We want to show that $c$ is the only limit point of the sequence. Let $c'$ be a limit point (not necessarily distinct from $c$ ), and let $\varepsilon > 0$ be arbitrary. Since the sequence converges to $c$ , there is some $N \geq m$ such that $|a_n - c| \leq \varepsilon/2$ for all $n \geq N$ . Fix this $N$ . Since $N \geq m$ and $c'$ is a limit point, there is some $n \geq N$ such that $|a_n - c'| \leq \varepsilon/2$ . Fix this $n$ . Since $n \geq N$ and the sequence converges to $c$ , we have $|a_n - c| \leq \varepsilon/2$ . Thus we have both $|a_n-c'| \leq \varepsilon/2$ and $|a_n - c| \leq \varepsilon/2$ . But this means $|c-c'| = |c-a_n+a_n-c'|\leq |c-a_n| + |a_n-c'| \leq \varepsilon$ . Since $\varepsilon > 0$ was arbitrary, we have just shown that $|c-c'| \leq \varepsilon$ for every $\varepsilon > 0$ . By Exercise 5.4.7 this means that $c=c'$ .

Solutions

Exercise 3.3.1

August 15, 2020 Issa Rice 14 Comments

Exercise statement

Show that the definition of equality in Definition 3.3.7 is reflexive, symmetric, and transitive. Also verify the substitution property: if $f, \tilde f: X \to Y$ and $g, \tilde g: Y \to Z$ are functions such that $f = \tilde f$ and $g = \tilde g$ , then $g \circ f = \tilde g \circ \tilde f$ .

Hints

None.

How to think about the exercise

This is a straightforward exercise. I do want to mention a couple of things though:

Definition 3.3.7 assumes that there is a notion of equality on elements of the range. If there was no such notion, then the statement $f(x)=g(x)$ that appears in the definition wouldn’t make sense.
The exercise further assumes that the notion of equality on elements of the range obey the properties of an equivalence relation (reflexive, symmetric, transitive) as well as the substitution axiom; these four properties are called the axioms of equality in Appendix A.7. There is no way to complete this exercise without assuming these properties.

Model solution

To show the definition is reflexive, let $f:X \to Y$ be a function. We must show $f = f$ . Let $x \in X$ . Then $f(x) = f(x)$ by reflexivity of equality on elements of $Y$ . Since $x$ was arbitrary, this shows that $f=f$ .

To show the definition is symmetric, let $f : X \to Y$ and $g : X \to Y$ be two functions with the same domain and range. Suppose $f=g$ . This means that $f(x)=g(x)$ for all $x \in X$ . To show that $g=f$ , we must show that $g(x)=f(x)$ for all $x \in X$ . So let $x \in X$ . Then since $f=g$ we have $f(x)=g(x)$ . By the symmetry of equality on the elements of $Y$ , we have $g(x)=f(x)$ . Since $x$ was arbitrary, this shows that $g=f$ .

To show the definition is transitive, let $f:X \to Y$ , $g:X\to Y$ , and $h:X\to Y$ be functions with the same domain and range. Suppose $f=g$ and $g=h$ . We must show that $f=h$ . So let $x \in X$ . Then we have $f(x)=g(x)$ since $f=g$ . We also have $g(x)=h(x)$ since $g=h$ . By transitivity of equality on elements of $Y$ , we have $f(x)=h(x)$ . Since $x$ was arbitrary, this shows that $f=h$ .

Now let $f, \tilde f: X \to Y$ and $g, \tilde g: Y \to Z$ be functions such that $f = \tilde f$ and $g = \tilde g$ . We want to show $g \circ f = \tilde g \circ \tilde f$ . Let $x \in X$ . Then since $f = \tilde f$ we have $f(x) = \tilde f(x)$ . By the substitution axiom of equality for elements of $Y$ (see Appendix A.7), we have $g(f(x)) = g(\tilde f(x))$ . Since $g = \tilde g$ , we have $g(\tilde f(x)) = \tilde g(\tilde f(x))$ . Now we have $(g\circ f)(x) = g(f(x)) = g(\tilde f(x)) = \tilde g(\tilde f(x)) = (\tilde g \circ \tilde f)(x)$ using symmetry and transitivity of equality on elements of $Z$ . Since $x \in X$ was arbitrary, this shows that $g \circ f = \tilde g \circ \tilde f$ as required.

Solutions

Exercise 6.3.4

August 1, 2020 Issa Rice 5 Comments

Exercise statement

Explain why Proposition 6.3.10 fails when $x > 1$ . In fact, show that the sequence $(x^n)_{n=1}^\infty$ diverges when $x > 1$ . Compare this with the argument in Example 1.2.3; can you now explain the flaws in the reasoning in that example?

Hints

Prove by contradiction and use the identity $(1/x)^n x^n = 1$ and the limit laws in Theorem 6.1.19.

How to think about the exercise

One thing that’s good to do when you’re trying to understand why a theorem fails in a particular case is to run through the proof of the theorem and locate the first step where the proof doesn’t make sense. This point comes early in the proof of Proposition 6.3.10 in the book. In particular, the sequence $(x^n)_{n=1}^\infty$ is not decreasing when $x > 1$ .

Model solution

Let $x > 1$ . Suppose for sake of contradiction that $\lim_{n\to\infty} x^n = L$ for some real number $L$ , i.e. that the sequence $(x^n)_{n=1}^\infty$ converges. Since $x > 1$ , we have $0 < 1/x < 1$ . Thus by Proposition 6.3.10 we have $\lim_{n\to\infty} (1/x)^n = 0$ . By the limit laws (Theorem 6.1.19(b)) we thus have $\lim_{n\to\infty} ((1/x)^n x^n) = \left(\lim_{n\to\infty} (1/x)^n\right) \left(\lim_{n\to\infty} x^n\right) = 0L = 0$ . But $(1/x)^n x^n = 1$ by Proposition 4.3.10(a), which we can use thanks to Proposition 5.6.3. Thus we can compute $\lim_{n\to\infty} ((1/x)^n x^n) = \lim_{n\to\infty} 1 = 1$ . This is a contradiction, and completes the proof.

Example 1.2.3 assumes that the limit $\lim_{n\to\infty} x^n$ exists for all real $x$ , which is false. If we restrict to $0 < x < 1$ then indeed we have $\lim_{n\to\infty} x^n = 0$ by Proposition 6.3.10.