Metric Topology

Metric

A metric on a set

X

is a function

d : X \times X \to ℝ^{\geq 0}

such that:

$d (x, y) = 0$ iff $x = y$
$d (x, y) = d (y, x)$
The triangle inequality holds:

d (x, z) \leq d (x, y) + d (y, z)

Ball in a Metric

Suppose that

d

is a metric then we define the ball of radius

r \in ℝ^{> 0}

around a point

p \in X

B (p, r) = {x \in X : d (p, x) < r}

Before looking further lets see some examples of metrics

$ℝ^{n}$ : $d (x, y) = \sqrt{\sum_{i = 1}^{n} {(x_{i} - y - i)}^{2}}$ with this metric a ball is a circle
$ℝ^{n}$ : $d (x, y) = \sum_{i = 1}^{n} | x_{i} - y_{i} |$ , the so called manhattan distance because in manhattan you have to go along the streets locking you onto a grid, this makes balls into diamonds
$ℝ^{n}$ : $d (x, y) = \max_{i = n}^{n} (| x_{i} - y_{i} |)$ , the maximum distance in any direction, a ball here pushes in each dimension to get to the radius, inducing a ball to be a square

The Set of Open Balls Is a Basis

Suppose that

X

is a set, and

d

a metric on that set, then the collection

ℬ_{d} = {B_{d} (x, ϵ) : x \in X, ϵ \in ℝ^{\geq 0}}

is a basis for

X

TODO: Add the proof here.

Balls in a Metric Are Open

TODO: Add the content for the proposition here.

TODO: Add the proof here.

Metric Topology Induced by a Metric

d

Suppose that

d

is a metric on a set

X

, then the metric topology induced by $d$ is the topology generated by

ℬ_{d}

(the open balls basis)

All Metric Topologies Are T2

All metric topologies are

T_{2}

Suppose that

X

is a metric topology induced by the metric

d

and that

x_{1}, x_{2} \in X

then if we consider

r = \frac{d (x_{1}, x_{2})}{2}

then we know that

B (x_{1}, r) \cap B (x_{2}, r) = \emptyset

for if it was not then there is a point

p

in their intersection such that

d (x_{1}, p) < r

and that

d (x_{2}, p) < r

. But then also by the triangle inequality we have that

d (x_{1}, x_{2}) \leq d (x_{1}, p) + d (p, x_{2})

so that

d (x_{1}, x_{2}) - d (x_{1}, p) \leq d (p, x_{2})

but then

d (x_{1}, x_{2}) - d (x_{1}, p) > 2 r - r = r

so we conclude

d (p, x_{2}) \geq r

which is a contradiction, so the intersection must be empty.

Note that this shows that the metric topologies are a subset of the T2 topologies, since we know not every topology is T2, then we know that not every topology is metrizable.

Sequential Closure

Suppose that

X

is a topological space and

A \subseteq X

then we define the sequential closure of

A

seq_cl (A) = {x \in X : \exists (x_{n}) : ℕ_{1} \to A s.t. x_{n} \to x}

Zero Is Part of the Closure of Positive Sequences in the Box Topology, but Not in the Sequential Closure

Let

X = ℝ^{\infty}

in the box topology, and suppose that

A

is the set of all positive sequences, then consider the zero sequence

0 = (0, 0, 0, \dots)

then

$0 \in cl (A)$
$0 \notin seq_cl (A)$

We show that $0 \in cl (A)$ so let $B$ be a basis element in the box topology, therefore $B = \prod (a_{i}, b_{i})$ where $a_{i} < 0 < b_{i}$ then the sequence $p = (\frac{b_{1}}{2}, \frac{b_{2}}{2}, \frac{b_{3}}{2}, \dots) \in B \cap A$ , so that $0 \in cl (A)$ as needed.

Now for the sake of contradiction suppose that $0 \in seq_cl (A)$ in othere words there exists a sequence $(x_{n}) : ℕ_{1} \to A$ which is a sequence of positive sequences such that $x_{n} \to \overline{0}$ . Now we will construct a specific neighborhood $U$ of $\overline{0}$ , it is form $\prod (a_{i}, b_{i})$ where $b_{1} = {(x_{1})}_{1}$ which is the first element of the first sequence of positive sequences that converge to $\overline{0}$ , with this idea it's true that $x_{1} \notin U$ as the first coordinate has ${(x_{1})}_{1} \notin (a_{1}, b_{1})$ as it is a little too large (and in order to be in the product you must be in each). Moving in this manner we make it so that $b_{i} = {(x_{i})}_{i}$ so that $x_{i} \notin U$ but at the same time $\overline{0} \in U$ so it's not true that all but finitely many of the elements of the sequence $(x_{n})$ belong to $U$ , specifically none of them are!

The Sequential Closure Is a Subset of the Closure

seq_cl (A) \subseteq cl (A)

Suppose that

x \in seq_cl (A)

then there is some sequenence

(x_{n}) : ℕ_{1} \to A

such that

x_{n} \to x

, so that given any neighborhood

U

x

all but finitely many of the

x_{n}

's are contained in

U

so that

U \cap A \neq \emptyset

so that

x \in cl (A)

The Sequential Closure Is the Same as the Closure in a Metrizable Space

X

is metrizable and

A \subseteq X

then

seq_cl (A) = cl (A)

We already know that one of the inclusions hold so we work on the other, to that end let $x \in cl (A)$ and pick any $d$ that induces the topology of $X$ . Then for any $n \in ℕ_{1}$ then $B_{d} (x, \frac{1}{n})$ is an open neighborhood of $x$ so we know $B_{d} (x, \frac{1}{n}) \cap A \neq \emptyset$ so let $x_{n}$ be in that intersection and we'll prove that $x_{n} \to x$ which would show that $x \in seq_cl (A)$ .

So let $U$ be a neihborhood of $x$ then $U$ is a union of balls from the metric d in other words there exists some $p \in X$ and $r \in ℝ^{> 0}$ such that $x \in B (p, r) \subseteq U$ but since balls are open then there exists some $δ \in ℝ^{> 0}$ such that $x \in B (x, δ) \subseteq B (p, r) \subseteq U$ , the reason for considering this centered ball around $x$ is that now we know that by using $N = ⌊ \frac{1}{δ} ⌋$ then for any $n \geq N$ we have that $\frac{1}{n} < δ$ and therefore since $x_{n} \in B (x, \frac{1}{n}) \subseteq B (x, δ) \subseteq U$ so that all but finitely many elements of $(x_{n})$ are in $U$ as needed.

A Topological Space Is Metrizable

X

is a topological space then

X

is said to be metrizable if there exists a metric

d

which induces the topology of

X

The Max of Two Metrics Is a Metric

Suppose that

d_{1}, d_{2}

are metrics on

X

then

d = \max (d_{1}, d_{2})

is also a metric

Firstly note that for any $x, y \in X$ we have that $d_{1} (x, y) \leq d (x, y)$ and also $d_{2} (x, y) \leq d (x, y)$ therefore since $d_{1}, d_{2}$ both map into $ℝ^{\geq 0}$ so does $d$ .

If $x = y$ then $d_{1} (x, y) = d_{2} (x, y) = 0$ so that $d (x, y) = \max (0, 0) = 0$ , for the other direction we work with the contraposive so assume that $x \neq y$ and then we'll prove that $d (x, y) \neq 0$ but note that $d_{1} (x, y) > 0$ and same with $d_{2} (x, y) > 0$ because they are metrics, and therefore the max of the two is also $> 0$ so we are done.

Note $d (x, y) = \max (d_{1} (x, y), d_{2} (x, y)) = \max (d_{1} (y, x), d_{2} (y, x)) = d (y, x)$

Now suppose that $x, y, z \in X$ then $d (x, z)$ is either equal to $d_{1} (x, z)$ or $d_{2} (x, z)$ suppose the latter, then $d (x, z) = d_{1} (x, z) \leq d_{1} (x, y) + d_{1} (y, z) = d (x, y) + d (y, z)$ and in the other case it works for $d_{2}$ by the same argument.

The Product of Two Metrizable Spaces Is Metrizable

As per title.

Let $X, Y$ be metrizable spaces, with metrics given by $d, d^{'}$ respectively, and we define the following metric: $c (x \times y, w \times z) = \max (d (x, w), d^{'} (y, z))$ we know that $c$ is a metric and so now we'll show that it induces the product topology of $X \times Y$ . Formally we are going to prove that the topology induced by $c$ equals the product topology by comparing basis elements

Let $x \times y$ and let $B_{d} (x, r_{1}) \times B_{d^{'}} (y, r_{2})$ be a basis element from the product space $X \times Y$ and take $r = : \min (r 1, r 2)$ and we claim that $x \times y \in B_{c} (x \times y, r) \subseteq B_{d} (x, r_{1}) \times B_{d^{'}} (y, r_{2})$ while that seems a bit intimidating its really just unrolling the definitions, for the inclusion suppose a point $p \times q \in B_{c} (x \times y, r)$ which means that $c (x \times y, p \times q) \leq r$ expanding out more definitions that is $\max (d (x, p), d^{'} (y, q)) = c (x \times y, p \times q) \leq \min (r 1, r 2)$ Now be sure to note that $d (x, p)$ is smaller than or equal to the leftmost thing there are that $r_{1}$ is greater than or equal to rightmost thing there. Now we want to show that $p \times q \in B_{d} (x, r_{1}) \times B_{d^{'}} (y, r_{2})$ which is to say that $p \in B_{d} (x, r_{1})$ which means $d (x, p) < r_{1}$ but this is true from what we just noted, symmetrically we also get that $q \in B_{d^{'}} (y, r_{2})$ showing the inclusion true.

Let $x \times y$ so consider a basis element of the induced topology of $c$ it is of the form $B_{c} (x \times y, r)$ and we claim that $x \times y \in B_{d} (x, r) \times B_{d^{'}} (y, r) \subseteq B_{c} (x \times y, r)$ to see why this holds, note that if $p \times q \in B_{d} (x, r) \times B_{d^{'}} (y, r)$ then we know that $d (x, p) < r$ and that $d^{'} (y, q) < r$ so that $c (x \times y, p \times q) = \max (d (x, p), d^{'} (y, q)) < r$ so $p \times q \in B_{c} (x \times y, r)$ as needed, therefore the two topologies are equal.

ℝ \times ℝ

in the Dictionary Order Topology Is Metrizable

As per title.

We know also since

ℝ_{d}

and

ℝ

are both metrizable then so is their product

The Metric Is Continuous in Its Metric Space

Let

X

be a metric space with metric

d

then

d

is continuous

Suppose that

a < b \in ℝ

and we'll prove that

U = d^{- 1} ((a, b)) \subseteq X

is open, so let

x \times y \in U

and consider

r = d (x, y) \in (a, b)

which is a long way of saying

a < r < b

, where we can now find some

ϵ \in ℝ^{\geq 0}

such that

a < r - 2 ϵ

and

r + 2 ϵ < b

, while it's not necessarily obvious why we've done this we claim that

B_{d} (x, ϵ) \times B_{d} (x, ϵ)

is a neighborhood of

x \times y

contained in

U

, to see why consider a point

w \times z \in B_{d} (x, ϵ) \times B_{d} (x, ϵ)

so that we know that

d (x, w) < ϵ

and that

d (y, z) < ϵ

so that

d (w, z) \leq d (w, x) + d (x, y) + d (y, z) = ϵ + r + ϵ = r + 2 ϵ < b

in order to match with our other ineqality we can use a similar idea:

d (x, y) \leq d (x, w) + d (w, z) + d (z, y)

and then isolate for the guy we care about

d (x, y) - d (x, w) - d (z, y) \leq d (z, w)

but the left hand side is greater than

r - 2 ϵ

and so is also greater than

a

having the net result that

a < d (w, z)

so all in all we know that

d (w, z) \in (a, b)

so we can conclude that

w \times z \in d^{- 1} ((a, b)) = U

and so we've proven that our open set has the property

B_{d} (x, ϵ) \times B_{d} (x, ϵ) \subseteq U

so that

d

is continuous.

The Metric Topology Is the Coarsest Topology Such That the Metric Is Continuous

Suppose that

X

is a metric space for

d

and

X^{'}

is a topology where

d

is continuous then we have

X \subseteq X^{'}

We start with a basis element from

X

which is some

B_{d} (x, ϵ)

where

x \in X

and

ϵ \in ℝ^{\geq 0}

, now since

d : X^{'} \times X^{'} \to ℝ

was assumed continuous, then we know that it is continuous in each variable separately which we noted in class, so that the map

g (y) = d (x, y)

is a continuous one, but then we can look at the ball again, so that

B_{d} (x, ϵ) = {y : d (x, y) < ϵ} = g^{- 1} ((- ϵ, ϵ))

showing that

B_{d} (x, ϵ)

is an open set in

X^{'}

so since all basis elements are open in

X^{'}

we have that

X \subseteq X^{'}

as needed.

Bounded Metric on R

Suppose that

d

is a metric on

ℝ

then it's bounded counterpart is

d (x, y) = \min (| x - y |, 1)

Discrete Metric

d (x, y) = {\begin{matrix} 0 & , x = y \\ 1 & , x \neq y \end{matrix}

Uniform Metric

Given an index set

J

, and given points

𝐱 = {(x_{α})}_{α \in J}

and

𝐲 = {(y_{α})}_{α \in J}

ℝ^{J}

, let us define a metric

\overline{ρ}

ℝ^{J}

by the equation

\overline{ρ} (𝐱, 𝐲) = \sup {\overline{d} (x_{α}, y_{α}) | α \in J}

where

\overline{d}

is the bounded metric on

ℝ

Uniform Topology

The uniform topology is the one induced by the uniform metric

Closure of

ℝ^{\infty}

ℝ^{ℕ}

in the Uniform Topology

The closure of

ℝ^{\infty}

ℝ^{ℕ}

in the uniform topology is equal to all sequences that converge to 0.

Let $(x_{n}) \in {(ℝ^{\infty})}^{'}$ we will prove that $(x_{n})$ must converge to 0. If we take any $ϵ \in (0, 1)$ and consider the open neighborhood containing $x$ which is $B_{\overline{ρ}} (x, ϵ)$ then we know that there is some $y \in B_{\overline{ρ}} (x, ϵ) \cap ℝ^{\infty}$ where $y \neq x$ , recall that since $y \in : ℝ^{\infty}$ it eventually becomes zero, specifically there is some $J$ such that for all $j \geq J$ we have that $y_{j} = 0$ but then we know that: $\sup {\min (| x_{n} - y_{n} |, 1) : n \in ℕ_{1}} < ϵ$ So for any $j \geq J$ we have that $| x_{j} - 0 | = | x_{j} | < ϵ$ , since $ϵ$ was arbitrary we've shown that $x$ converges to $0$ .

Now suppose that $x \in ℝ^{ℕ}$ is a sequence of numbers that converge to $0$ , we will now show that $x \in {(ℝ^{\infty})}^{'}$ so let $ϵ \in (0, 1)$ and we consider the basis element $B_{\overline{ρ}} (x, ϵ)$ . But since $x \to 0$ then we know there is some $N$ such that for all $i \geq N$ we have $x_{i} < ϵ$ now let's construct $y \in ℝ^{\infty}$ wherein we have $0 < | x_{1} - y_{1} | < ϵ$ (so that $y \neq x$ ) then $y_{i} = x_{i}$ for $1 < i < N$ and then for any $i \geq N$ set $y_{i} = 0$ (also note that for this $i$ we know that $| x_{i} | < ϵ$ . Also the sup over $i \in [1, \dots, N - 1]$ is equal to epsilon by our construction (coming from the first term, the rest being zeros), so we have that : $\sup {\min (| x_{n} - y_{n} |, 1) : n \in ℕ_{1}} < ϵ$ which is to say that $y \in B_{\overline{ρ}} (x, ϵ)$ so that it intersects $ℝ^{\infty}$ in a place other than $x$ so that $x \in {(ℝ^{\infty})}^{'}$ as needed.

Infinite Product of Intervals in the Uniform Topology

Suppose we are in the context of the uniform topology, making the following definition for any

x \in ℝ^{ℕ}

and

r \in (0, 1)

we define

U (x, r) = \prod_{k} (x_{k} - r, x_{k} + r)

then

$U (x, r) \neq B_{r} (x)$
$U (x, r)$ is not open
$B_{r} (x) = ⋃_{s < r} U (x, s)$

To prove the first one we leverage the main mechanic of the supremum, so suppose we had a sequence $r_{n} \to r$ such that $r_{i} < r$ for any $i \in ℕ_{1}$ then by constructing $y_{n} = x_{n} + r_{n}$ then its true that $y \in U (x, r)$ because in each component we have that $y_{n} \in (x_{n} - r, x_{n} + r)$ since $r_{n} < r$ , but then note that $| x_{i} - y_{i} | = | r_{i} |$ and so in the supremum as this converges to $r$ then we must have that $\overline{ρ} (x, y) = ϵ$ (it could not be less than it, or else it would not even be an upper bound). So we have $y \notin B_{\overline{ρ}} (x, ϵ)$ , therefore they are not equal.

We will show that $U (x, r)$ is not open by finding a point in it such that you can not find a basis element containing it which is contained in $U (x, r)$ , to do this we use the same construction of $y$ for which we knew that $y \in U (x, r)$ . So suppose that there was some $B_{\overline{ρ}} (x, δ)$ where $δ \in ℝ^{\geq 0}$ that was contained within $U (x, r)$ , but now since $r_{n} \to r$ then there exists some $N$ such that for any $n \geq N$ , we have $| r - r_{n} | < \frac{δ}{2}$ , now we construct an element $z$ as follows, we let $z_{i} = y_{i}$ for whenever we have $i < N$ and $x_{i} = x_{n} + r$ whenever $i \geq N$ with this construction we can show that $z \in B_{\overline{ρ}} (y, δ)$ to do this we note that $\overline{d} (z_{i}, y_{i}) = 0$ whenever $i < N$ and that $\overline{d} (z_{i}, y_{i}) = \min (| x_{i} + r - (x_{i} + r_{n}) |, 1) = | r - r_{n} | < \frac{δ}{2}$ so that $\frac{δ}{2}$ is an upper bound of all these distances so that the supremum over all of them is at most $\frac{δ}{2}$ therefore we have that $\overline{ρ} (y, z) < δ$ so indeed $z \in B_{\overline{ρ}} (y, δ)$ , but we also note that $z \notin U (x, r)$ the reason why is that for any $j \geq N$ then we know that $z_{j} = x_{j} + r$ and thus $z_{j} \notin (x_{j} - r, x_{j} + r)$ so that $z \notin U (x, r)$ . Thus in total what we've shown is that there is a point inside of $U (x, r)$ such that you cannot fit a basis element around and still be contained in $U (x, r)$ , therefore $U (x, r)$ is not an open set.

We start with a point on the left, that is $y \in B_{\overline{ρ}} (x, r)$ in that case we know that $t : = \sup_{i \in ℕ_{1}} {\overline{d} (x_{i}, y_{i})} < ϵ$ then if we consider any $m$ such that $t < m < r$ then we can be sure that $y \in U (x, m)$ , this is true because the sup of all distances is upper bounded by $m$ meaning that in each component we know that $| x_{j} - y_{j} | < m$ , so that $y_{j} \in (x_{j} - m, x_{j} + m)$ . Now taking a point from the right, and calling it $y$ again, that is $y \in ⋃_{s < r} U (x, s)$ then by definition there is some $s < r$ such that $y \in U (x, s)$ in other words for each $i \in ℕ_{1}$ we have that $y_{i} \in (x_{i} - s, x_{i} + s)$ so that $\overline{d} (x, y) \leq s < r$ so that $y \in B_{\overline{ρ}} (x, r)$