# A. Let (X,\ d) be a metric space. Define a

A. Let $$\displaystyle{\left({X},\ {d}\right)}$$ be a metric space. Define a diameter of a subset A of X and then the diameter of an open ball with center at $$\displaystyle{x}_{{{0}}}$$ and radios $$\displaystyle{r}{>}{0}$$
B. Use (A) to show that every convergence sequence in $$\displaystyle{\left({X},\ {d}\right)}$$ is bounded.

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enlacamig

Step 1
A) Let $$\displaystyle{\left({X},\ {d}\right)}$$ be a metric space A be a subset of x.
We define diameter of A to be
Sup $$\displaystyle{\left\lbrace{d},\ {\left({x},\ {y}\right)}{\mid}{x},\ {y}\in{A}\right\rbrace}$$
Let $$\displaystyle\cup$$ be an open boll with center at $$\displaystyle{x}_{{{0}}}$$ and radious $$\displaystyle{r}{>}{0}$$
Then, diameter of $$\displaystyle\cup={2}{r}$$
We choose x, y to be diametrically opposite two points whose midpoint is $$\displaystyle{x}_{{{0}}}$$
Then $$\displaystyle{d}{\left({x},\ {y}\right)}$$ atfains 2r.
Also, for any $$\displaystyle{x},\ {y}\in\cup$$
$$\displaystyle{d}{\left({x},\ {y}\right)}\le{d}{\left({x},\ {x}_{{{0}}}\right)}+{d}{\left({x}_{{{0}}},\ {y}\right)}$$
$$\displaystyle\le{r}+{r}={2}{r}$$
$$\displaystyle\therefore\ {d}{i}{a}{m}{\left(\cup\right)}={2}{r}$$
Step 2
B) Let $$\displaystyle{\left({x}_{{{n}}}\right)}_{{{n}\not\in{{{N}}}}}$$ be a squence which is convergent in $$\displaystyle{\left({x},\ {d}\right)}$$
We assume that, it converges to $$\displaystyle{x}\in x$$
Choose $$\displaystyle\sum={1}$$
Then $$\displaystyle\exists{N}\not\in{{{N}}}$$
s.t $$\displaystyle\forall{h}\geq\ {d}{\left({x}_{{{n}}},\ {x}\right)}{ < }{1}$$
We take
$N1=Sup\begin{cases}d(x,\ x_{1}), & d(x,\ x_{2})\\ \dots, & d(x,\ x_{n-1}),1\} \end{cases}$
Then for any $$\displaystyle{x}_{{{n}}}$$ in the sep
$$\displaystyle{d}{\left({x}_{{{n}}},\ {x}\right)}{ < }{1}$$
So, $$\displaystyle{\left({x}_{{{n}}}\right)}_{{{n}\not\in{{{N}}}}}$$ is bounded

###### Not exactly what you’re looking for?
Cassandra Ramirez
Step 1
Consider the discrete metric d on a set X:
$$\displaystyle{d}{\left({x},\ {y}\right)}={\left\lbrace\begin{array}{cc} {0}&\ {\quad\text{if}\quad}{x}={y}\\{1}&\ {\quad\text{if}\quad}{x}\ne{y}\end{array}\right.}$$
Consider the ball of radius $$\displaystyle{r}={\frac{{{1}}}{{{2}}}}$$ centered at x
Then $$\displaystyle{B}{\left({x},\ {r}\right)}={\left\lbrace{x}\right\rbrace}$$
Now by definition, diam
$$\displaystyle{A}=\text{sup}{\left\lbrace{d}{\left({a},{b}\right)}:{a},{b}\in{A}\right\rbrace}$$
Applying it to our case where
$$\displaystyle{A}={B}{\left({x},{r}\right)}$$
we have diameter of A equal to 0
nick1337

A fairly common example that isn't completely trivial is the British Rail metric on $$\mathbb{R}$$, where
d(x, y)=|x|+|y|
(or 0 for x=y), so B(x, r)
Has diameter 0 for x<r,
diameter r for x<r<2x and diameter
2(r-x)
for
r>2x