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 \(\mathscr{T}\).

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 \(X\).

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.

Proof:

Will be added later.