every subspace of a normed space of finite dimension is closed
Let (V,∥⋅∥) be a normed vector space, and S⊂V a finite dimensional subspace
. Then S is closed.
Proof
Let a∈ˉS and choose a sequence {an} with an∈S such that
an converges
to a. Then {an} is a Cauchy sequence
in V and
is also a Cauchy sequence in S. Since a finite dimensional normed
space is a Banach space
, S is complete
, so {an} converges to an
element of S. Since limits in a normed space are unique, that limit
must be a, so a∈S.
Example
The result depends on the field being the real or complex numbers. Suppose the V=Q×R, viewed as a vector space over Q and S=Q×Q is the finite dimensional subspace. Then clearly (1,√2) is in V and is a limit point of S which is not in S. So S is not closed.
Example
On the other hand, there is an example where Q is the underlying
field and we can still show a finite dimensional subspace is closed. Suppose
that V=Qn, the set of n-tuples of rational numbers, viewed
as vector space over Q. Then if S is a finite dimensional subspace
it must be that S={x|Ax=0} for some matrix A.
That is, S is the inverse image of the closed set
{0}.
Since the map x→Ax is continuous
, it follows that S is a closed set.
Title | every subspace of a normed space of finite dimension![]() |
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Canonical name | EverySubspaceOfANormedSpaceOfFiniteDimensionIsClosed |
Date of creation | 2013-03-22 14:56:28 |
Last modified on | 2013-03-22 14:56:28 |
Owner | Mathprof (13753) |
Last modified by | Mathprof (13753) |
Numerical id | 12 |
Author | Mathprof (13753) |
Entry type | Theorem |
Classification | msc 54E52 |
Classification | msc 15A03 |
Classification | msc 46B99 |