Solutions to Problem set 4

Chapter 2: Problem 10

This is the `discrete metric' on a set. Certainly $d:X\times X\longrightarrow [0,\infty)$ is well defined and $d(x,y)=0$ iff $x=y.$ Symmetry, $d(x,y)=d(y,x),$ is immediate from the definition and the triangle inequality

$$d(x,y)\le d(x,z)+d(z,y)\ \forall\ x,y,x\in X$$ (10)

follows from the fact that the right hand side is always equal to $0,$ $1$ or $2$ and the LHS is 0 or $1$ and if the LHS vanishes then $x=y=z$ and the RHS also vanishes.

All subsets are open, since if $E\subset X$ and $p\in E$ then $x\in X$ and $d(x,p)<1$ implies $x=p$ and hence $x\in E.$ Since the complements of open sets are closed it follows that all subsets are closed. The only compact subsets are finite. Indeed if $E\subset X$ is compact then the open balls of radius $1$ with centers in $E$ cover $E$ and each contains only one point of $E$ so the existence of a finite subcover implies that $E$ itself is finite.

Chapter 2: Problem 12

We are to show that $K=\{1/n;n\in\mathbb{N}\}\cup\{0\}$ is compact as a subset of $\mathbb{R}$ directly from the definition of compactness. So, let $U_a,$ $a\in A,$ be an open cover of $K.$ It follows that $0\in U_{a_0}$ for some $a_0\in A.$ But since $U_{a_0}$ is open it contains some ball of radius $1/n$ around $0.$ Thus all the points $1/m\in U_{a_0}$ for $m>n.$ For each $m\le n$ we can find some $a_m\in A$ such that $1/m\in U_{a_m},$ since the $U_a$ cover $K.$ Thus we have found a finite subcover

$$K\subset U_{a_0}\cup U_{a_1}\cup\dots\cup U_{a_n}$$ (11)

and if follows that $K$ is compact.

Chapter 2: Problem 16

Here $\mathbb{Q}$ is the metric space, with $d(p,q)=\vert p-q\vert,$ the `usual' metric. Set

$$E=\{p\in\mathbb{Q};2<p^2<3\}.$$ (12)

Suppose $x$ is a limit point of $E$ as a subset of the rationals. Then we know that $(x-\epsilon ,x+\epsilon )\cap E$ is infinite for each $\epsilon >0.$ Regarding $x$ as a real number it follows that $x\in[2^{\frac12},3^{\frac12}].$ Since we know the end points are not rational and by assumption $x\in\mathbb{Q}$ it follows that $x\in E.$ Thus $E$ is closed. Certainly $E$ is bounded since $p\in E$ implies $\vert p\vert<3.$

To see that $E$ is not compact, recall that if it were compact as a subset of $\mathbb{Q}$ it would be compact as a subset of $\mathbb{R}$ by Theorem 2.33. Since it is not closed as a subset of $\mathbb{R}$ it cannot be compact. Alternatively, for a direct proof of non-compactness, take the open cover given by the open sets $\{x\in\mathbb{Q};\vert p-2^{\frac12}\vert>1/n\}.$ This can have no finite subcover since $E$ contains points arbitrarily close to the real point $\sqrt2.$

Yes $E$ is open in $\mathbb{Q}$ since it is of the form $(\sqrt2,\sqrt3)\cap\mathbb{Q}$ where $G=(\sqrt2,\sqrt3)\subset\mathbb{R}$ is open, so Theorem 2.30 applies.

Chapter 2: Problem 22

We need to show that the set of rational points, $\mathbb{Q}^k$ is dense in $\mathbb{R}^k.$ We can use the fact that $\mathbb{Q}\subset\mathbb{R}$ is dense. Thus, given $\epsilon >0$ and $x=(x_1,\dots,x_k)\in\mathbb{R}^k$ there exists $p_l\in\mathbb{Q}$ such that $\vert x_l-p_l\vert\le \epsilon/k$ for each $l=1,\dots,k.$ Thus, as points in $\mathbb{R}^k,$

$$\vert x-p\vert\le \sum\limits_{l=1}^k\vert x_l-p_l\vert<\epsilon .$$ (13)

This shows that $\mathbb{R}^k$ is separable since we know that $\mathbb{Q}^k$ is countable.

Chapter 2: Problem 23

We are to show that a given separable metric space, $X,$ has a countable base. The hint is to choose a countable dense subset $E\in X$ and then to consider the collection, $\mathcal B,$ of all open subsets of $X$ of the form $B(x,1/n)$ where $x\in E$ and $n\in\mathbb{N}.$ This is a countable union, over $\mathbb{N},$ of countable sets so is countable. Now, we need to show that this is a base. So, suppose $U\subset X$ is a given open set. If $x\in U$ then for some $m=m_x>0,$ $B(x,1/m)\subset U,$ since it is open. Also, by the density of $E$ in $X$ there exists some $e_x\in E$ with $\vert x-e_x\vert<1/2m$ But then $y\in B(e_x,1/2m)$ implies $d(x,y)<d(x,e_x)+d(e_x,y)<1/2m+1/2m=1/m.$ Thus $B(e_x,1/2m)\subset U.$ It follows that

$$U=\bigcup_{x\in U}B(e_x,1/2m_x).$$ (14)

Thus $U$ is written as a union of the elements of $\mathcal B$ which is therefore an open base.

Chapter 2: Problem 25

We wish to show that a given compact metric space $K$ has a countable base. As the hint says, for each $n\in \mathbb{N}$ consider the balls of radius $1/n$ around each of the points of $K:$

$$K\subset \bigcup_{x\in K}B(x,1/n)$$ (15)

since each $x\in K$ is in one of these balls at least. Now the compactness of $K$ implies that there is a finite subcover, that is there is a finite subset $C_n\subset K,$ for each $n,$ such that

$$K\subset \bigcup_{p\in C_n}B(p,1/n).$$ (16)

Now, set $E=\bigcup_{n\in\mathbb{N}}C_n.$ This is countable, being a countable union of finite sets. Now it follows that $C$ is dense in $K.$ Indeed given $x\in K$ and $\epsilon >0$ there exists $n\in \mathbb{N}$ with $1/n<\epsilon$ and from (16) a point in $p\in C_n\subset C$ with $\vert x-p\vert<1/n<\epsilon.$ Thus $K=\overline C$ and it follows that $K$ is separable; from #23 it follows that $K$ has a countable open base.

Alternatively one can see directly that the $B(p,1/n),$ $p\in C_n,$ form an open base.