3
$\begingroup$

Is it possible to take the one-point compactification of a linear space and get again a linear space? Like if $X$ is a normed space add $*$ to it and define $|x-*| = 1$?

$\endgroup$
5
$\begingroup$

This can't be achieved if you want the normed space structure to be compatible with the topology in the sense that the norm is continuous. Since otherwise, the norm function, now having compact domain, must achieve a maximum, which is impossible due to homogeneity.

In case what you are after is just 'compactifying' vector spaces (with no underlying topology), then you'll have to be more precise about what compact should mean. In any case, you can't take a vector space (in the presence of enough scalars), add a point to it, and extend the vector space structure to a vector space. The reason is that for every scalar $\alpha\ne 0 $ in the ground field, $x\mapsto \alpha x$ is injective. So if you add one new element, you must add at least $|\mathbb F|-1 $ other elements, where $\mathbb F $ is the ground field.

$\endgroup$
  • $\begingroup$ Does your argument with homogeneity still apply if the underlying field only contains finitely many scalars? $\endgroup$ – snailspace Jun 26 '13 at 7:03
2
$\begingroup$

If $|x - *| = 1$ for all $x \in X$, then $$ |x| \le |x - *| + |*| = 2 $$ for all $x$. If there exists $y \in X$ with $|y| > 0$, then there exists an integer $n$ with $n|y| = |ny| > 2$. Therefore, all $x \in X$ must satisfy $|x| = 0$.

$\endgroup$
0
$\begingroup$

Suppose $\hat{\Bbb R}$ is a copmpatification of $\Bbb R$ and the normal norm on $\Bbb R$ can be extended to $\hat{\Bbb R}$. Then $\Bbb R$ is a topological subgroup of the topological group $\hat{\Bbb R}$. This means that $\Bbb R$ is precompact. Precompactness of $\Bbb R$ means any sequence in $\Bbb R$ must have a Cauchy subsequence. $\{n^2\}$ has not a Cauchy subsequence. A contradiction.

See also: this post and Alex Ravsky's comments in this thread.

If anything is wrong let me know. I'll fix or delete this post.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.