Tutorial 1 Workings

1. Show that the limits of Cauchy sequences in a complete metric space are unique.

Let $(X,d)$ be a complete metric space and $(x_n)_{n=1}^{\infty}$ be a Cauchy sequence in $X$. Suppose that the sequence converges to $L_1$ and $L_2$ in $X$. We claim that $L_1=L_2$.

Let $\varepsilon>0$ be arbitrary. There then exists some positive integers $N_1, N_2$ such that

$$d(x_n,L_1)<\varepsilon/2 \text{ whenever } n \geqslant N_1,$$$$d(x_n,L_2)<\varepsilon/2 \text{ whenever } n \geqslant N_2.$$

Let $N=\max\{N_1,N_2\}$. It follows that whenever $n \geqslant N$, we have that

$$d(L_1,L_2) \leqslant d(L_1,x_n)+d(x_n,L_2)<\varepsilon.$$

Since the choice of $\varepsilon$ is arbitrary, this implies that $d(L_1,L_2)=0$, which means that $L_1=L_2$.

2. Prove the reverse triangle inequality:

$$|d(x,z)-d(z,y)| \leqslant d(x,y)$$

for all $x,y,z$ in a metric space $(X,d)$.

We first prove that $d(x,z)-d(z,y) \leqslant d(x,y)$. Indeed, it follows from the triangle inequality that $d(x,z) \leqslant d(x,y)+d(y,z)$.

Next, we prove that $-d(x,y) \leqslant d(x,z)-d(z,y)$, which is equivalent to $d(z,y)-d(x,z) \leqslant d(x,y)$. This is again just the triangle inequality: $d(z,y) \leqslant d(z,x)+d(x,y)$.

3. Give an example of a sequence $\{x_n\} \in \mathbb{R}^\infty$ such that $x_n \to 0$, but $\{x_n\} \notin l^p$ for any $1 \leqslant p < \infty$.

One such sequence $x=(x_n)_{n=1}^{\infty}$ is given by

$$x_n=\frac{1}{\ln(n+1)}.$$

The divergence of $\Vert x \Vert_p$ when $p \in [1,\infty)$ can be seen through limit comparison test with $\sum_{n=1}^{\infty} \frac{1}{n}$.

4. Let $(X,d)$ and $(Y,\rho)$ be metric spaces and suppose that $f : X \to Y$ is uniformly continuous. Prove that under $f$, the image of every Cauchy sequence is a Cauchy sequence.

Let $(x_n)_{n=1}^{\infty}$ be a Cauchy sequence in $(X,d)$. Let $\delta>0$ be arbitrary, then there exists some positive integer $N$ such that

$$d(x_n,x_m)<\delta \text{ whenever } n,m \geqslant N.$$

We map the sequence via $f$ and obtain a sequence $(f(x_n))_{n=1}^{\infty}$ in $(Y,\rho)$. Since $f$ is uniformly continuous (at an arbitrary point $b \in X$), for every $\varepsilon > 0$, there exists some $\delta>0$ such that (for points $a \in X$ satisfying...)

$$d(a,b)<\delta \implies \rho(f(a),f(b))<\varepsilon.$$

Since $(x_n)_{n=1}^{\infty}$ is Cauchy, for every $\varepsilon>0$, we can find a positive integer $N$ such that $d(x_n,x_m)<\delta$ whenever $n,m \geqslant N$ that gives $\rho(f(x_n),f(x_m))<\varepsilon$ whenever $n,m \geqslant N$.

Therefore, $(f(x_n))_{n=1}^{\infty}$ is also Cauchy.

5. Suppose $W$ is a dense subset of a metric space $X$ and $f$ is a uniformly continuous function from $W$ into some complete metric space $Y$. Prove that $f$ has a unique continuous extension to $X$. That is, there exists a unique continuous function $F : X \to Y$ such that $F(w)=f(w)$ for all $w \in W$.

This immediately follows from Exercise 43.2 of Munkres' Topology. Note that the denseness of $W$ implies that the closure of $W$ is precisely $X$ itself.

6. Given two non-empty subsets $A,B$ of a metric space $(X,d)$, their distance is defined as

$$D(A,B)=\inf_{a \in A,b \in B}d(a,b).$$

Consider the power set of $X$ and the function $D$. Which of the axioms of a metric space does this pair satisfy?

It satisfies positivity, since $D$ is induced by $d$, which is a metric that thus satisfies positivity. If $d(a,b) \geqslant 0$ for all $a,b \in X$ (i.e. 0 is a lower bound of $d$), then $\inf_{a \in A,b \in B}d(a,b) \geqslant 0$ for all $A,B \in \mathcal{P}(X)$ (the greatest lower bound of $d$ is indeed greater than one of its lower bounds, 0).

It also satisfies symmetry. This immediately follows from the symmetry of the metric $d$ that induces $D$.

It does not satisfy definiteness. If $A=B$, then $\inf_{a \in A,b \in B}d(a,b)=0$ since we can just take common points that have zero distance between them. However, the converse is not true: if $\inf_{a \in A,b \in B}d(a,b)=0$, it means that there are common points in $A$ and $B$, i.e. the intersection $A \cap B$ is not empty, but this does not necessarily mean that $A=B$.

It does not satisfy triangle inequality either. Take $X=\mathbb{R}$, $D$ be induced by the standard metric on the real line, $A$ be the open interval $(2,3)$, $B$ be the open interval $(3,4)$ and $C$ be the open interval $(5,6)$. It follows that $D(A,C)=2$, but $D(A,B)+D(B,C)=0+1 < 2$.