Topology/Euclidean Spaces

From Wikibooks, open books for an open world
Jump to: navigation, search

Euclidean space is the space in which everyone is most familiar. In Euclidean k-space, the distance between any two points is

d(x,y) = \sqrt {\sum^k_{i=1} {(x_i - y_i)^2}}

where k is the dimension of the Euclidean space. Since the Euclidean k-space as a metric on it, it is also a topological space.


Definition: A sequence of real numbers \{s_n\} is said to converge to the real number s provided for each \epsilon > 0 there exists a number N such that n > N implies \mid s_n - s \mid  < \epsilon.

Definition: A sequence \{s_n\} of real numbers is called a Cauchy sequence if for each \epsilon > 0 there exists a number N such that m,n > N \Rightarrow \mid s_n -s_m \mid  < \epsilon.

Lemma 1[edit]

Convergent sequences are Cauchy Sequences.

Proof : Suppose that lim \; s_n = s.


\mid s_n - s_m \mid =  \mid s_n - s + s - s_m \mid \leq \mid s_n - s\mid + \mid s - s_m \mid

Let \epsilon > 0. Then \exists N such that

n > N \Rightarrow \mid s_n - s \mid < \frac {\epsilon}{2}


m > N \Rightarrow \mid s_m - s \mid < \frac {\epsilon}{2}


m,n > N \Rightarrow \mid s_n - s_m \mid \leq \mid s_n - s\mid + \mid s - s_m \mid < \frac {\epsilon}{2} + \frac{\epsilon}{2} = \epsilon

Hence, \{s_n\} is a Cauchy sequence.

Theorem 1[edit]

Convergent sequences are bounded.

Proof: Let \{s_n\} be a convergent sequence and let lim \; s_n = s. From the definition of convergence and letting \epsilon = 1, we can find N \in \mathbb{N} such that

n > N \Rightarrow \mid s_n - s \mid < 1

From the triangle inequality;

n > N \Rightarrow \mid s_n \mid < \mid s \mid + 1

Let M = max \{\mid s_n \mid + 1, \mid s_1 \mid , \mid s_2 \mid , ... , \mid s_N \mid \}.


\mid s_n \mid \leq M

for all n \in \mathbb{N}. Thus \{s_n\} is a bounded sequence.

Theorem 2[edit]

In a complete space, a sequence is a convergent sequence if and only if it is a Cauchy sequence.


\Rightarrow Convergent sequences are Cauchy sequences. See Lemma 1.

\Leftarrow Consider a Cauchy sequence \{s_n\}. Since Cauchy sequences are bounded, the only thing to show is:

\liminf s_n = \limsup s_n

Let \epsilon > 0. Since \{s_n\} is a Cauchy sequence, \exists N such that

m,n > N \Rightarrow \mid s_n - s_m \mid  <  \epsilon

So, s_n < s_m + \epsilon for all m,n > N. This shows that s_m + \epsilon is and upper bound for \{s_n : n > N\} and hence v_N = \sup \{s_n : n > N\} \leq s_m + \epsilon for all  m > N. Also v_N - \epsilonis a lower bound for \{s_m : m > N\}. Therefore v_n - \epsilon \leq \inf \{s_m : m > N\} = u_N. Now:

\limsup s_n \leq v_N \leq u_N + \epsilon \leq \liminf s_n + \epsilon

Since this holds for all \epsilon > 0, \limsup s_n \leq \liminf s_n. The opposite inequality always holds and now we have established the theorem.

Note: The preceding proof assumes that the image space is \mathbb{R}. Without this assumption, we will need more machinery to prove this.


It is possible to have more than one metric prescribed to an image space. It is often better to talk about metric spaces in a more general sense. A sequence \{s_n\} in a metric space (S,d) converges to s in S if \lim _{n \rightarrow\infty} d(s_n,s) = 0. A sequence is called Cauchy if for each \epsilon > 0 there exists an N such that:

m,n > N \Rightarrow d(s_m,s_n) < \epsilon


The metric space (S,d) is called complete if every Cauchy sequence in S converges to some element in S.

Theorem 3[edit]

Let X be a complete metric and Y be a subspace of X. Then Y is a complete metric space if and only if Y is a closed subset of X.

Proof: \Leftarrow Suppose Y is a closed subset of X. Let \{s_n\} be a Cauchy sequence in Y.

Then \{s_n\} is also a Cauchy sequence in X. Since X is complete, \{s_n\} converges to a point s in X. However, Y is a closed subset of X so Y is also complete.

\Rightarrow Left as an exercise.


1. Let x,y \in \mathbb{R}^k. Let d_1(x,y) = max \mid x_i - y_i \mid where \{i = 1,2,...,k\} and d_2(x,y) = \sum_i^k \mid x_i - y_i \mid. Show:

a.) d_1 and d_2 are metrics for \mathbb{R}^k.

b.) d_1 and d_2 form a complete metric space.

2. Show that every open set in \mathbb{R} is the disjoint union of a finite or infinite sequence of open intervals.

3. Complete the proof for theorem 3.

4.Consider: Let X and Y be metric spaces.

a.) A mapping T : X \rightarrow Y is said to be a Lipschitz mapping provided that there is some non-negative number c called a Lipschitz constant for the mapping such that:

d[T(x),T(y)] \leq cd(x,y) \; \forall x,y \in X

b.) A Lipschitz mapping T : X\rightarrow Y that has a Lipschitz constant less than 1 is called a contraction.

Suppose that f : \mathbb{R} \rightarrow\mathbb{R} and g :\mathbb{R} \rightarrow\mathbb{R} are both Lipschitz. Is the product of these functions Lipschitz?