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

William Curry 2021-12-21 Answered
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
Answered 2021-12-22 Author has 178 answers

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

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Cassandra Ramirez
Answered 2021-12-23 Author has 136 answers
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
0
nick1337
Answered 2021-12-28 Author has 9467 answers

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

0

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