# Exercise 7.2.17 Introduction to Real Analysis by Jiri Lebl

Let $$(X, d)$$ be a metric space.

a) Prove that for every $$x \in X$$, there either exists a $$\delta> 0$$ such that $$B(x, \delta) = \{x\}$$, or $$B(x, \delta)$$ is infinite for every $$\delta >0$$.

If $$X$$ is discrete, then the first statement is true. If $$X$$ is connected, then the second statement is true.

b) Find an explicit example of $$(X, d)$$, $$X$$ infinite, where for every $$\delta >0$$ and every $$x \in X$$, the ball $$B(x, \delta)$$ is finite.

Let $$X = \mathbb{N}$$. Then, since $$\delta$$ is finite, $$B(x, \delta)$$ is finite.

c) Find an explicit example of $$(X, d)$$ where for every $$\delta >0$$ and every $$x \in X$$, the ball $$B(x, \delta)$$ is countably infinite.

Let $$X = \mathbb{Q}$$. Rational number is countably infinite in any interval (?), so the statement is true.

d) Prove that if $$X$$ is uncountable, then there exists an $$x \in X$$ and a $$\delta>0$$ such that $$B(x, \delta)$$ is uncountable.

If $$B(x, \delta)$$ is countable for all $$x \in X$$, every $$x$$ in the ball can be mapped to $$\mathbb{Q}$$ (?), so $$X$$ must be countable, which is a contradiction.

All answers might be wrong or incomplete. I appreciate if you give some help for each question.

Your answer to a) is wrong . You cannot take a specific metric space to prove it. Suppose $$B(x,\delta)$$ is finite for every $$\delta >0$$. In particular $$B(x,1)$$ is finite. Let $$x_1,x_2,..,x_n$$ be the distinct points other than $$x$$ itself in this ball. Now take $$\delta=\frac 1 2 \min\{d(x,x_1),d(x,x_2),...,d(x,x_n)\}$$. Now you can easily check that $$B(x, \delta)=\{x\}$$.
For d) let $$x$$ be any point and assume that $$B(x,\delta)$$ is at most countable for every $$\delta>0$$ . Then $$X =\bigcup_{n=1}^{\infty} B(x,n)$$ is at most countable which is a contradiction.
• (d) would need to only consider only rational $\delta$ to be a countable collection of at most countable sets. – Matthew Daly Jan 5 at 23:26
• @MatthewDaly Why rational $\delta$? Why not integer $\delta$? – Kavi Rama Murthy Jan 5 at 23:27
• Integer would work as well, but it should be more clear than using $n$ as a variable that that is what you intend. – Matthew Daly Jan 5 at 23:29
• @MatthewDaly Conventionally, the notation $_{n=1}^{\infty}$ implies $n$ varying over $\mathbb{N}$., no? – WoolierThanThou Jan 5 at 23:39