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.

  • $\begingroup$ (d) would need to only consider only rational $\delta$ to be a countable collection of at most countable sets. $\endgroup$ – Matthew Daly Jan 5 at 23:26
  • $\begingroup$ @MatthewDaly Why rational $\delta$? Why not integer $\delta$? $\endgroup$ – Kavi Rama Murthy Jan 5 at 23:27
  • $\begingroup$ Integer would work as well, but it should be more clear than using $n$ as a variable that that is what you intend. $\endgroup$ – Matthew Daly Jan 5 at 23:29
  • $\begingroup$ @MatthewDaly Conventionally, the notation $_{n=1}^{\infty}$ implies $n$ varying over $\mathbb{N}$., no? $\endgroup$ – WoolierThanThou Jan 5 at 23:39

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