Topology/Euclidean Spaces

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

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   is said to converge to the real number s provided for each   there exists a number   such that   implies  .

Definition: A sequence   of real numbers is called a Cauchy sequence if for each   there exists a number   such that        .

Lemma 1Edit

Convergent sequences are Cauchy Sequences.

Proof : Suppose that  .



Let  . Then   such that






Hence,   is a Cauchy sequence.

Theorem 1Edit

Convergent sequences are bounded.

Proof: Let   be a convergent sequence and let  . From the definition of convergence and letting  , we can find N   such that


From the triangle inequality;


Let  .



for all  . Thus   is a bounded sequence.

Theorem 2Edit

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


  Convergent sequences are Cauchy sequences. See Lemma 1.

  Consider a Cauchy sequence  . Since Cauchy sequences are bounded, the only thing to show is:


Let  . Since   is a Cauchy sequence,   such that


So,   for all  . This shows that   is and upper bound for   and hence   for all  . Also  is a lower bound for  . Therefore  . Now:


Since this holds for all  ,  . The opposite inequality always holds and now we have established the theorem.

Note: The preceding proof assumes that the image space is  . 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   in a metric space   converges to s in S if  . A sequence is called Cauchy if for each   there exists an   such that:



The metric space   is called complete if every Cauchy sequence in   converges to some element in  .

Theorem 3Edit

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

Proof:   Suppose   is a closed subset of  . Let   be a Cauchy sequence in  .

Then   is also a Cauchy sequence in  . Since   is complete,   converges to a point   in  . However,   is a closed subset of   so   is also complete.

  Left as an exercise.


1. Let  . Let   where   and  . Show:

a.)   and   are metrics for  .

b.)   and   form a complete metric space.

2. Show that every open set in   is the disjoint union of a finite or infinite sequence of open intervals.

3. Complete the proof for theorem 3.

4.Consider: Let   and   be metric spaces.

a.) A mapping   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:


b.) A Lipschitz mapping   that has a Lipschitz constant less than 1 is called a contraction.

Suppose that   and   are both Lipschitz. Is the product of these functions Lipschitz?