5.7 KiB
Topological notions
Definition 1: let
(X,d)
be a metric space and letx_0 \in X
andr > 0
, the following may be defined
- open ball:
B(x_0, r) = \{x \in X \;|\; d(x,x_0) < r\}
,- closed ball:
\tilde B(x_0,r) = \{x \in X \;|\; d(x,x_0) \leq r\}
,- 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 ifX \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 letx_0 \in X
, thenB(x_0, \varepsilon)
is an $\varepsilon$-neighbourhood ofx_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 ofx_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 letM \subset X
, a pointx_0 \in M
is an interior point ofM
ifM
is a neighbourhood ofx_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 letM \subset X
, the interior ofM
, denoted byM^\circ
is the set of all interior points ofM
.
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 ofX
, then
\empty \in \mathscr{T} \land X \in \mathscr{T}
,- the union of a collection of sets in
\mathscr{T}
is itself a set in\mathscr{T}
,- 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 letT: X \to Y
be a map.T
is continuous inx_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 allx_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 mappingT: X \to Y
is continuous if and only if the inverse image of any open subset ofY
is an open subset ofX
.
??? 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 pointx_0 \in X
is an accumulation point ofM
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 ofM
is the closure ofM
, denoted by\overline M
.
Therefore, \overline M
is the smallest closed set containing M
.
Definition 10: let
(X,d)
be a metric space and letM
be a subset ofX
. The setM
is dense inX
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 ifX
contains a countable subsetM
that is dense inX
.
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.