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

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

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

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

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

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

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

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

Richard B. Melrose 2004-05-18