In this section we will characterize the normed linear spaces of finite dimension.
Any two norms on a finite dimensional normed linear space induce the same topology; in other words, any two norms are equivalent.
Proof. Let be a finite dimensional normed linear space. Let be a basis for . Then any can be written with each . Define a Euclidean norm on by We want to show that the norm on and determine the same topology, i.e. that any -open set contains a -open set and vice versa. Let Thus, and so any -open -ball contains a -open -ball.
Now we want to show that any -open ball contains a -open ball. Define by . This function is continuous in the Euclidean topology (use the inequality above). The unit sphere in , , is compact, hence attains a minimum .
So for any with we have . This extends to all of by properties of the norm. If , then , and . Therefore, by multiplying both sides of the inequality by , we get the desired inequality for all in . Thus, any any -open ball contains a -open ball.
Thus, we have shown that any norm on a finite dimensional normed linear space is equivalent to the Euclidean norm. In fact, is homeomorphic to for some .
A direct consequence of this is the following proposition.
If is a finite dimensional subspace of a normed linear space , then is closed.
Proof. Including the above proof, we know the following about : Any norm of is equivalent to the Euclidean norm on ; Every Cauchy sequence in is Cauchy in ; is a complete metric space, thus every Cauchy sequence converges; Convergence in implies convergence in ; is a complete subspace of the metric space , thus is closed.
It follows from this proposition that every finite dimensional normed linear space is complete. Next question to ask: Is every subspace of a normed linear space closed? And if we were to find a non-closed subspace of a normed linear space where would we have to look?
Let denote the subspace of consisting of those sequences which have only finitely many non-zero terms. For any let The sequence of these sequences, , is a subset of that converges to Note that is in but not in ; therefore is not closed. Indeed,
The last property we need to show is that finite dimensional gives us
closed and bounded equals
compact and vice versa. We first need a lemma.
Let be a closed, proper subspace of a normed linear space . Then for any , there exists a with and for all .
Proof. Let . The distance from to , , is positive (else you can find a sequence of elements in converging to , which would contradict that is a closed subspace). Let be a positive real number. There exists such that . Let . Then . We want to show that with the right choice of we get the inequality we want. Because, for any , we just need to choose so that .
A normed linear space is finite dimensional if and only if every closed and bounded subset is compact.
Proof. Let be a finite dimensional normed linear space. Let be a closed and bounded subset. By Proposition I.62, closed and bounded in the norm implies closed and bounded in the Euclidean norm. So, the Heine-Borel Theorem applies and is compact.
Suppose every closed, bounded subset of is compact. By way of contradiction, suppose that is infinite dimensional. We'll construct a bounded sequence with no convergent subsequence to get the contradiction.
Let such that . Let . By Proposition I.63, is closed and by Lemma I.65, there exists an such that and . Let . As before, is closed and there exists an such that , , and . Continue forever, which you can do because is infinite dimensional. In this way, we get a bounded infinite sequence where for all , and when . Therefore, has no convergent subsequence, which contradicts the assumption that every closed, bounded subset is compact.
The closed unit ball in an infinite-dimensional, normed linear space is not compact.
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