notes/docs/mathematics/functional-analysis/metric-spaces/topological-notions.md

Topological notions

Definition 1: let (X,d) be a metric space and let x_0 \in X and r > 0, the following may be defined

1. open ball: B(x_0, r) = \{x \in X \;|\; d(x,x_0) < r\},
2. closed ball: \tilde B(x_0,r) = \{x \in X \;|\; d(x,x_0) \leq r\},
3. sphere: S(x_0,r) = \{x \in X \;|\; d(x,x_0) = r\}.

In all three cases x_0 can be thought of as the center and r as the radius.

Definition 2: a subset M \subset X of a metric space (X,d) is open if \forall x_0 \in M \exists r > 0: B(x_0,r) \subset M.

M is closed if X \backslash M is open.

Therefore, one may observe that an open ball is an open set and a closed ball is a closed set.

Neigbourhoods

Definition 3: let (X,d) be a metric space and let x_0 \in X, then B(x_0, \varepsilon) is an $\varepsilon$-neighbourhood of x_0 for some \varepsilon > 0.

Using definition 3 we may define the following.

Definition 4: a neighbourhood of x_0 is a set that contains an $\varepsilon$-neighbourhood of x_0 for some \varepsilon > 0.

Therefore x_0 is an element of each of its neighbourhoods and if N is a neighbourhood of x_0 and N \subset M, then M is also a neighbourhood of x_0.

Definition 5: let (X,d) be a metric space and let M \subset X, a point x_0 \in M is an interior point of M if M is a neighbourhood of x_0.

One may think of an interior point of a subset as a point that lies within the interior of M.

Definition 6: let (X,d) be a metric space and let M \subset X, the interior of M, denoted by M^\circ is the set of all interior points of M.

One may observe that M^\circ is open and is the largest open set contained in M.

Lemma 1: let (X,d) be a metric space and let \mathscr{T} be the set of all open subsets of X, then

1. \empty \in \mathscr{T} \land X \in \mathscr{T},
2. the union of a collection of sets in \mathscr{T} is itself a set in \mathscr{T},
3. the intersection of a finite collection of sets in \mathscr{T} is a set in \mathscr{T}.

??? note "Proof:"

Statement 1 follows by noting that $\empty$ is open since $\empty$ has no elements and $X$ is open. 

For statement 2 we have that for any point $x$ of the union $U$ of open sets belongs to at least one of these sets $M$ and $M$ contains a ball $B$ about $x$. Then $B \subset U$, by the definition of a union. 

For statement 3 we have that if $y$ is any point of the intersection of open sets $M_1, \dots, M_n$ with $n \in \mathbb{N}$ then each $M_j$ contains a ball about $y$ and the smallest of these balls is contained in that intersection. 

From statements 1 and 3 from lemma 1 we may define a topological space (X,\mathscr{T}) to be a set X and a collection \mathscr{T} of subsets of X such that \mathscr{T} satisfies the axioms 1 and 3. The set \mathscr{T} is a topology for X, and it follows that a metric space is a topological space.

Continuity

Definition 7: let (X,d) and (Y,\tilde d) be metric spaces and let T: X \to Y be a map. T is continuous in x_0 \in X if

\forall \varepsilon > 0 \exists \delta > 0 \forall x \in X: \quad d(x,x_0) < \delta \implies \tilde d \big(T(x), T(x_0) \big) < \varepsilon.

A mapping T is continuous if it is continuous in all x_0 \in X.

Continuous mappings can be characterized in terms of open sets as follows.

Theorem 1: let (X,d) and (Y,\tilde d) be metric spaces, a mapping T: X \to Y is continuous if and only if the inverse image of any open subset of Y is an open subset of X.

??? note "Proof:"

Suppose that $T$ is continuous. Let $S \subset Y$ be open and $S_0$ the inverse image of $S$. If $S_0 = \empty$, it is open. Let $S_0 = \empty$. For any $x \in S_0$ let $y_0 = T(x_0)$. Since $S$ is open, it contains an $\varepsilon$-neighbourhood $N$ of $y_0$. Since $T$ is continuous, $x_0$ has a $\delta$-neighbourhood $N_0$ which is mapped into $N$. Since, $N \subset S$ we have $N_0 \subset S_0$ so that $S_0$ is open because $x_0 \in S_0$ is arbitrary.

Suppose that the inverse image of every open set in $Y$ is an open set in $X$. Then for every $x_0 \in X$ and any $\varepsilon$-neighbourhood $N$ of $T(x_0)$, the inverse image $N_0$ of $N$ is open, since $N$ is open, and $N_0$ contains $x_0$. Hence, $N_0$ also contains a $\delta$-neighbourhood of $x_0$, which is mapped into $N$ because $N_0$ is mapped into $N$. Consequently, $T$ is continuous at $x_0$. Since $x_0 \in X$ was chosen arbitrary, $T$ is continuous.

Accumulation points

Definition 8: let M \subset X be a subset of a metric space (X,d). A point x_0 \in X is an accumulation point of M if

\forall \varepsilon > 0 \exists y \in M \backslash {x_0}: d(x_0,y) < \varepsilon.

An accumulation point of a subset M is also sometimes called a limit point of M. Implying the nature of these points.

Definition 9: the set consisting of all points of M and all accumulation points of M is the closure of M, denoted by \overline M.

Therefore, \overline M is the smallest closed set containing M.

Definition 10: let (X,d) be a metric space and let M be a subset of X. The set M is dense in X if \overline M = X.

Hence if M is dense in X, then every ball in X, no matter how small, will contain points of M.

Definition 11: a metric space (X,d) is separable if X contains a countable subset M that is dense in X.

For example the real line \mathbb{R} is separable, since the set \mathbb{Q} of all rational numbers is countable and is dense in \mathbb{R}.

Furthermore, l^\infty is not separable while l^p is indeed separable.

??? note "Proof:"

Will be added later.