Let $X $ be a set. A sequence in $X $ will be denoted by
$$\pare{X^{(n)}}^ \infty_ {n = m} $$
where $m$ is the initial term of the sequence, $n$ is the term of the sequence you are considering.
Let $(x_n)^ \infty _{n=m}$ be a sequence of real numbers, and let $x $ be another real number. Then
$$(x_n)^ \infty _ {n=1} \text{ converges to }x \triangleq \limu{ n }{ \infty }d(x_n, x) = 0$$
Equivalently,
$$(x_n)^ \infty _ {n=1} \text{ converges to }x \triangleq \text{Fix }\epsilon > 0, \exists N > 0\text{ s.t. }\forall n > N , d(x^{(n)}, x_0) < \epsilon $$
A metric space is any space $X$ which has a concept of distance $d(x, y)$ - and this distance should behave in a reasonable manner. More precisely,
A metric space $(X, d)$ is a space $X$ ob objects (called points), together with a distance function or metric $d: X \times X \to [0, +\infty)$, which associates to each pair $x, y$ of points in $X $ a non-negative real number $d(x, y) \geq 0$. Furthermore, the metric must satisfy the following four axioms:
- For any $x \in X $, we have $d(x, x) = 0$.
- (Positivity) For any distinct $x, y \in X$, we have $d(x,y) > 0$.
- (Symmetry) For any $x, y \in X $, we have $d(x, y) = d(y, x)$.
- (Triangle inequality) For any $x, y, z \in X$, we have $d(x, z) \leq d(x, y) + d(y, z)$.
In many cases, it will be clear what the metric $d$ is, and we shall abbreviate $(X, d)$ as just $X$.
The conditions (a) and (b) can be rephrased as follows:
for any $x, y \in X $ we have $d(x, y) = 0 $ if and only if $x= y $.
$$d(x,y) = \begin{cases} 0 &x = y \br 1 &x \neq y \br \end{cases}$$
Let $X $ be any set, and let $d_{ \text{disc}} $ be the discrete metric on $X $.
Let $(x^{(n)})^ \infty _{n = m} $ be a sequence of points in $X $, and let $x $ be a point in $X $.
Then $(x^{(n)}) ^ \infty _{n = m}$ converges to $x $ with respect to the discrete metric $d_{ \text{disc}} $ if and only if there exists an $N \geq m $ such that $x^{(n)} = x $ for all $n \geq N $.
Let $X $ be any set, $d$ be the discrete metric.
$(X, d) $ is always a metric space.
$\R^n = \set{ x = (x_1, _2, …, _{ n }) : x_1, x_2, …, x_{ n } \in \R } $.
Metrics on $\R^n$.
- $l^1$ metric, or Taxi-cab metric.
- $l^2$ metric, or Euclidean / Standard metric.
- $l^ \infty $ metric, or sup norm metric.
Let $n \geq 1 $, and let $\R^n $ be as before. $d_{l^ 1} $, the so-called taxicab metric (or {l^ 1} metric), is defined by
$$d_{l^1} ((x_1, _2, …, _{ n }), (y_1, (y_2, …, (y_{ n })) \triangleq \abs{ x_1 - y_1 } + … + \abs{ x_n - y_n } = $$
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Let $n \geq 2 $, and let $\R^n $ be as before.
$$d_{l^2} ((x_1, _2, …, _{ n }), (y_1, (y_2, …, (y_{ n })) \triangleq \sqrt[ ]{ \sum_{ i=1 }^{ n } \abs{ x_i - y_i }^2 }$$
Let $n \geq 1 $, and let $\R^n $ be as before. But now we use a different metric $d_l^ \infty $, the so-called sup norm metric(or $l^ \infty $ metric), defined by
$$d_{l^ \infty} ((x_1, _2, …, _{ n }), (y_1, (y_2, …, (y_{ n })) \triangleq \supr{ \abs{ x_i - y_i } : 1 \leq i \leq n }$$
Thus for instance, if $n =2 $, then $d_{l^ \infty} ((1, 6), (4, 2)) = \supr{ 3, 4 } = 4 $. The space $(\R^n, d_{l^ \infty})$ is also a metric space, and is related to the $l^2 $ metric by the inequalities
$\frac{1}{\sqrt{n}}d_{l^ 2} (x, y) \leq d_{l^ \infty} (x, y) \leq d_{l^2} (x, y)$ for all $x, y $.
$x^{(n)} = (\frac{ 1 }{ n }, \frac{ 1 }{ n }).$ Does $(x^{(n)})^ \infty_ {n = 1}$ converge in $d_{l^1} $?
Guess that $x_0 = (0, 0) $ is a great candidate for a limit.
$d(x^{(n)}, x_0) = \sum_{ i = 1 }^{ 2 } \abs{ \frac{ 1 }{ n } - 0 } = \frac{2}{n} $
$\lim d(x^{(n)}, x_0) = 0 $
Hence, $x^{(n)} = (\frac{ 1 }{ n }, \frac{ 1 }{ n })$ converges.