9 KiB
Completeness
Definition 1: a sequence
(x_n)_{n \in \mathbb{N}}
in a metric space(X,d)
is a Cauchy sequence if
\forall \varepsilon > 0 \exists N \in \mathbb{N} \forall n,m > N: \quad d(x_n, x_m) < \varepsilon.
A convergent sequence (x_n)_{n \in \mathbb{N}}
in a metric space (X,d)
is always a Cauchy sequence since
\forall \varepsilon > 0 \exists N \in \mathbb{N}: \quad d(x_n, x) < \frac{\varepsilon}{2},
for all n > N
. By axiom 4 of the definition of a metric space we have for m, n > N
d(x_m, x_n) \leq d(x_m, x) + d(x, x_n) < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon,
showing that (x_n)
is Cauchy.
Definition 2: a metric space
(X,d)
is complete if every Cauchy sequence inX
is convergent.
Therefore, in a complete metric space every Cauchy sequence is a convergent sequence.
Proposition 1: let
M \subset X
be a nonempty subset of a metric space(X,d)
and let\overline M
be the closure ofM
, then
x \in \overline M \iff \exists (x_n)_{n \in \mathbb{N}} \text{ in } M: x_n \to x
,M \text{ is closed } \iff M = \overline M
.
??? note "Proof:"
To prove statement 1, let $x \in \overline M$. If $x \notin M$ then $x$ is an accumulation point of $M$. Hence, for each $n \in \mathbb{N}$ the ball $B(x,\frac{1}{n})$ contains an $x_n \in M$ and $x_n \to x$ since $\frac{1}{n} \to 0$ as $n \to \infty$. Conversely, if $(x_n)_{n \in \mathbb{N}}$ is in $M$ and $x_n \to x$, then $x \in M$ or every neighbourhood of $x$ contains points $x_n \neq x$, so that $x$ is an accumulation point of $M$. Hence $x \in \overline M$.
Statement 2 follows from statement 1.
We have that the following statement is equivalent to statement 2: x_n \in M: x_n \to x \implies x \in M
.
Proposition 2: let
M \subset X
be a subset of a complete metric space(X,d)
, then
M \text{ is complete} \iff M \text{ is a closed subset of } X
??? note "Proof:"
Let $M$ be complete, by proposition 1 statement 1 we have that
$$
\forall x \in \overline M \exists (x_n)_{n \in \mathbb{N}} \text{ in } M: x_n \to x.
$$
Since $(x_n)$ is Cauchy and $M$ is complete, $x_n$ converges in $M$ with the limit being unique by statement 1 in [lemma 1](). Hence, $x \in M$ which proves that $M$ is closed because $x \in \overline M$ has been chosen arbitrary.
Conversely, let $M$ be closed and $(x_n)$ Cauchy in $M$. Then $x_n \to x \in X$ which implies that $x \in \overline M$ by statement 1 in proposition 1, and $x \in M$ since $M = \overline M$ by assumption. Hence, the arbitrary Cauchy sequence $(x_n)$ converges in $M$.
Proposition 3: let
T: X \to Y
be a map from a metric space(X,d)
to a metric space(Y,\tilde d)
, then
T \text{ is continuous in } x_0 \in X \iff x_n \to x_0 \implies T(x_n) \to T(x_0),
for any sequence
(x_n)_{n \in \mathbb{N}}
inX
asn \to \infty
.
??? note "Proof:"
Suppose $T$ is continuous at $x_0$, then for a given $\varepsilon > 0$ there is a $\delta > 0$ such that
$$
\forall \varepsilon > 0 \exists \delta > 0: \quad d(x, x_0) < \delta \implies \tilde d(Tx, Tx_0) < \varepsilon.
$$
Let $x_n \to x_0$ then
$$
\exists N \in \mathbb{N} \forall n > N: \quad d(x_n, x_0) < \delta.
$$
Hence,
$$
\forall n > N: \tilde d(Tx_n, Tx_0) < \varepsilon.
$$
Which means that $T(x_n) \to T(x_0)$.
Conversely, suppose that $x_n \to x_0 \implies T(x_n) \to T(x_0)$ and $T$ is not continuous. Then
$$
\exists \varepsilon > 0: \forall \delta > 0 \exists x \neq x_0: \quad d(x, x_0) < \delta \quad \text{ however } \quad \tilde d(Tx, Tx_0) \geq \varepsilon,
$$
in particular, for $\delta = \frac{1}{n}$ there is a $x_n$ satisfying
$$
d(x_n, x_0) < \frac{1}{n} \quad \text{ however } \quad \tilde d(Tx_n, Tx_0) \geq \varepsilon,
$$
Clearly $x_n \to x_0$ but $(Tx_n)$ does not converge to $Tx_0$ which contradicts $Tx_n \to Tx_0$.
Completeness proofs
To show that a metric space (X,d)
is complete, one has to show that every Cauchy sequence in (X,d)
has a limit in X
. This depends explicitly on the metric on X
.
The steps in a completeness proof are as follows
- take an arbitrary Cauchy sequence
(x_n)_{n \in \mathbb{N}}
in(X,d)
, - construct for this sequence a candidate limit
x
, - prove that
x \in X
, - prove that
x_n \to x
with respect to metricd
.
Proposition 4: the Euclidean space
\mathbb{R}^n
withn \in \mathbb{N}
and the metricd
defined by
d(x,y) = \sqrt{\sum_{j=1}^n \big(x(j) - y(j) \big)^2},
for all
x,y \in \mathbb{R}^n
is complete.
??? note "Proof:"
Let $(x_m)_{m \in \mathbb{N}}$ be a Cauchy sequence in $(\mathbb{R}^n, d)$, then we have
$$
\forall \varepsilon > 0 \exists N \in \mathbb{N}: \forall m, k > N: d(x_m, x_k) = \sqrt{\sum_{j=1}^n \big(x_m(j) - x_k(j) \big)^2} < \varepsilon,
$$
obtains for all $j \in \mathbb{N}$: $|x_m(j) - x_k(j)| < \varepsilon$.
Which shows that $(x_m(j))_{m \in \mathbb{N}}$ is a Cauchy sequence in $\mathbb{R}$. Suppose that it converged by $x_m(j) \to x(j)$ as $(m \to \infty)$ then $x \in \mathbb{R}^n$ since $x = \big(x(1), \dots, x(n)\big)$.
Thus for $(k \to \infty)$ we have
$$
d(x_m, x) < \varepsilon \implies x_m \to x,
$$
which implies that $\mathbb{R}^n$ is complete.
A similar proof exists for the completeness of the Unitary space \mathbb{C}^n
.
Proposition 5: the space
C([a,b])
of all real-valued continuous functions on a closed interval[a,b]
witha<b \in \mathbb{R}
with the metricd
defined by
d(x,y) = \max_{t \in [a,b]} |x(t) - y(t)|,
for all
x, y \in C
is complete.
??? note "Proof:"
Let $(x_n)_{n \in \mathbb{N}}$ be a Cauchy sequence in $(C,d)$, then we have
$$
\forall \varepsilon > 0 \exists N \in \mathbb{N}: \forall n, m > N: d(x_n, x_m) = \max_{t \in [a,b]} |x_n(t) - x_m(t)| < \varepsilon,
$$
obtains for all $t \in [a,b]$: $|x_n(t) - x_m(t)| < \varepsilon$.
Which shows that $(x_m(t))_{m \in \mathbb{N}}$ for fixed $t \in [a,b]$ is a Cauchy sequence in $\mathbb{R}$. Since $\mathbb{R}$ is complete the sequence converges; $x_m(t) \to x(t)$ as $m \to \infty$.
Thus, for $m \to \infty$ we have
$$
d(x_n, x) = \max_{t \in [a,b]} | x_n(t) - x(t) | < \varepsilon,
$$
hence $\forall t \in [a,b]: | x_n(t) - x(t) < \varepsilon$, obtaining convergence to $x_n \to x$ as $n \to \infty$ and $x \in C$ which implies that $C$ is complete.
While C
with a metric d
defined by
d(x,y) = \int_a^b |x(t) - y(t)| dt,
for all x,y \in C
is incomplete.
??? note "Proof:"
Will be added later.
Proposition 6: the space
l^p
withp \geq 1
and the metricd
defined by
d(x,y) = \Big(\sum_{j \in \mathbb{N}} | x(j) - y(j) |^p\Big)^\frac{1}{p},
for all
x,y \in l^p
is complete.
??? note "Proof:"
Let $(x_n)_{n \in \mathbb{N}}$ be a Cauchy sequence in $(l^p,d)$, then we have
$$
\forall \varepsilon > 0 \exists N \in \mathbb{N}: n, m > N: d(x_n, x_m) = \Big(\sum_{j \in \mathbb{N}} |x_n(j) - x_m(j)|^p\Big)^\frac{1}{p} < \varepsilon,
$$
obtains for all $j \in \mathbb{N}$: $|x_n(j) - x_m(j)| <\varepsilon$.
Which shows that $(x_m(j))_{m \in \mathbb{N}}$ for fixed $j \in \mathbb{N}$ is a Cauchy sequence in $\mathbb{C}$. Since $\mathbb{C}$ is complete the sequence converges; $x_m(j) \to x(j)$ as $m \to \infty$.
Thus, for $m \to \infty$ we have
$$
d(x_n, x) = \Big(\sum_{j \in \mathbb{N}} |x_n(j) - x(j)|^p\Big)^\frac{1}{p} < \varepsilon,
$$
implies that $x_n - x \in l^p$ and $x = x_n - (x_n - x) \in l^p \implies x \in l^p$ and $x_n \to x$ as $n \to \infty$ which implies that $l^p$ is complete.
Proposition 7: the space
l^\infty
with the metricd
defined by
d(x,y) = \sup_{j \in \mathbb{N}} | x(j) - y(j) |,
for all
x,y \in l^\infty
is complete.
??? note "Proof:"
Let $(x_n)_{n \in \mathbb{N}}$ be a Cauchy sequence in $(l^\infty,d)$, then we have
$$
\forall \varepsilon > 0 \exists N \in \mathbb{N}: n, m > N: d(x_n, x_m) = \sup_{j \in \mathbb{N}} | x_n(j) - x_m(j) | < \varepsilon,
$$
obtains for all $j \in \mathbb{N}$: $|x_n(j) - x_m(j)| <\varepsilon$.
Which shows that $(x_m(j))_{m \in \mathbb{N}}$ for fixed $j \in \mathbb{N}$ is a Cauchy sequence in $\mathbb{C}$. Since $\mathbb{C}$ is complete the sequence converges; $x_m(j) \to x(j)$ as $m \to \infty$.
Thus, for $m \to \infty$ we have
$$
d(x_n, x) = \sup_{j \in \mathbb{N}} | x_n(j) - x(j) | < \varepsilon \implies |x_n(j) = x(j)| < \varepsilon.
$$
Since $x_n \in l^\infty$ there exists a $k_n \in \mathbb{R}: |x_n(j)| \leq k_n$ for all $j \in \mathbb{N}$. Hence
$$
|x(j)| \leq |x(j) - x_n(j)| + |x_n(j)| < \varepsilon + k_n,
$$
for all $j \in \mathbb{N}$ which implies that $x \in l^\infty$ and $x_n \to x$ as $n \to \infty$ obtaining that $ l^\infty$ is complete.