Separable Metric Space: Local Compactness and Completeness
Shozo Mori
Or is a complete separable metric space locally compact?
Thank you very much.
- Shozo
José Carlos Santos
> Is a locally compact separable metric space complete?
No. Consider an open interval (a,b) in R, with its usual distance.
> Or is a complete separable metric space locally compact?
Again, no. Consider the space of all absolutely convergent series
of real numbers with the distance
d(sum_n a_n,sum_n b_n) = sum_n |a_n - b_n|.
Best regards,
Jose Carlos Santos
David C. Ullrich
Good example, except you didn't say what you meant.
sum a_n is a _number_; you've defined the distance
between two reals, and in an ambiguous way since
a given real can be written as sum a_n in many different ways.
You meant he should consider the space of all absolutlely
summable _sequences_, with the metric
d((a_n), (b_n)) = sum_n |a_n - b_n|.
>Best regards,
>
>Jose Carlos Santos
David C. Ullrich
"Understanding Godel isn't about following his formal proof.
That would make a mockery of everything Godel was up to."
(John Jones, "My talk about Godel to the post-grads."
in sci.logic.)
Shozo Mori
Thank you very much for the responses, which I appreciate.
Another question: Is the (a ?) completion of a locally compact, separable metric space locally compact and separable?
Probably the answer is yes.
If so, I might say any locally compact, separable metric space is "almost" complete. Do you agree?
Or, the answer may be "No," and Jose's counter example would be also a counter example of the above question. Which do you agree with?
By the way, the fact that a complete separable metric space (Polish space ?) is "more general" than a locally compact separable metric space is "more or less" obvious, when we consider a vector space, i.e., a Euclidean space vs. C([0 1]).
Do you agree?
Again thank you very much.
- Shozo
José Carlos Santos
> Thank you very much for the responses, which I appreciate.
>
> Another question: Is the (a ?) completion
It is "a" completion, not "the" completion.
> of a locally compact,separable metric space locally compact
I don't know.
> and separable?
Of course, Because if A is a dense subset of B and B is a dense subset
of C, then A is a dense subset of C.
> Probably the answer is yes.
>
> If so, I might say any locally compact, separable metric space
> is "almost" complete. Do you agree?
Actually, *any* metric space is "almost complete", in the sense that
every metric space is isometric to a dense subspace of a complete metric
space.
David C. Ullrich
<jcsa...@fc.up.pt> wrote:
>On 12-11-2009 17:13, Shozo Mori wrote:
>
>> Thank you very much for the responses, which I appreciate.
>>
>> Another question: Is the (a ?) completion
>
>It is "a" completion, not "the" completion.
??? Of course that's literally true, but any two completions
of a metric space are isometric, so it's "the" completion,
up to the strongest possible relevant sort of isomorphism.
>> of a locally compact,separable metric space locally compact
>
>I don't know.
>
>> and separable?
>
>Of course, Because if A is a dense subset of B and B is a dense subset
>of C, then A is a dense subset of C.
>
>> Probably the answer is yes.
>>
>> If so, I might say any locally compact, separable metric space
>> is "almost" complete. Do you agree?
>
>Actually, *any* metric space is "almost complete", in the sense that
>every metric space is isometric to a dense subspace of a complete metric
>space.
>
>Best regards,
>
>Jose Carlos Santos
David C. Ullrich
José Carlos Santos
>> On 12-11-2009 17:13, Shozo Mori wrote:
>>
>>> Thank you very much for the responses, which I appreciate.
>>>
>>> Another question: Is the (a ?) completion
>> It is "a" completion, not "the" completion.
>
> ??? Of course that's literally true, but any two completions
> of a metric space are isometric, so it's "the" completion,
> up to the strongest possible relevant sort of isomorphism.
Right. :-( For a moment I was thinking about another type of completion,
namely about completion for a metrizable topological space. In general
there are several.
David C. Ullrich
wrote:
>Jose and David,
>
>Thank you very much for the responses, which I appreciate.
>
>Another question: Is the (a ?) completion of a locally compact, separable metric space locally compact and separable?
>
>Probably the answer is yes.
Of course it's yes for the "separable". The following is I think a
counterexample for "locally compact". Probably stupid, there must
be much simpler examples:
Say X is the space of all sequences a = (a_1, a_2, ...) of reals such
that ||a|| = sum |a_n| < infinity; define
d(a,b) = sum |a_j - b_j| = ||a-b||.
So X is a Banach space: a complete normed vector space.
Let H_1, H_2, ... be the sequence of finite-dimensional
subspaces of X: say a is in H_n if a is in X and a_j = 0 for
all j > n.
Choose a strictly in increasing sequence of positive numbers r_n
that tends to 1.
Let E be the union of the sets
E_n = {a in H_n : r_n <= ||a|| <= r_{n+1}}
Then it seems clear to me that E is locally compact, since any point
has a neighborhood isometric to a closed subset of some euclidean
space.
But the completion of E is the same as the closure of E in X,
which is the union of E and the unit sphere S = {a in X : ||a||=1}
This is not locally compact.