Is a colimit of Banach space topological vector space?
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question. 1
Let $X$ be a locally compact Hausdorff space, and
$$
mathcalC(X)= f : X rightarrow mathbbC mid f mathrmis continuous
$$
$$
mathcalC_c(X) = f in mathcalC(X) mid mathrmsuppf mathrm is compact
$$
and if $K subset X$ is compact,
$$
mathcalC(X,K) = f in mathcalC(X) mid mathrmsuppf subset K
$$
Define the topology of $mathcalC_c(X)$ as $varinjlim_K mathcalC(X,K)$.
($varinjlim_K mathcalC(X,K)$ means the colimt in $mathbfTop$ of Banach spaces $mathcalC_c(K)$)
Then, is $varinjlim_K mathcalC(X,K)$ topological vector space?
How to prove?
question.2
Especially when $X= mathbbR^n$, is $varinjlim_K mathcalC(X,K)$ topological vector space?
(I'm sorry my English is broken...I would be glad if you could help me.)
general-topology functional-analysis limits-colimits
add a comment |Â
up vote
0
down vote
favorite
question. 1
Let $X$ be a locally compact Hausdorff space, and
$$
mathcalC(X)= f : X rightarrow mathbbC mid f mathrmis continuous
$$
$$
mathcalC_c(X) = f in mathcalC(X) mid mathrmsuppf mathrm is compact
$$
and if $K subset X$ is compact,
$$
mathcalC(X,K) = f in mathcalC(X) mid mathrmsuppf subset K
$$
Define the topology of $mathcalC_c(X)$ as $varinjlim_K mathcalC(X,K)$.
($varinjlim_K mathcalC(X,K)$ means the colimt in $mathbfTop$ of Banach spaces $mathcalC_c(K)$)
Then, is $varinjlim_K mathcalC(X,K)$ topological vector space?
How to prove?
question.2
Especially when $X= mathbbR^n$, is $varinjlim_K mathcalC(X,K)$ topological vector space?
(I'm sorry my English is broken...I would be glad if you could help me.)
general-topology functional-analysis limits-colimits
It depends on the category in weich you take the colimit. In the category TVS of topological vector spaces the answer is trivially yes, but in the category of topological spaces it is probably no.
â Jochen
yesterday
Thank you for your reply. Does TVS has the filtered colimit?
â A. I
yesterday
Perhaps math.stackexchange.com/q/1432904 is useful?
â Paul Frost
yesterday
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
question. 1
Let $X$ be a locally compact Hausdorff space, and
$$
mathcalC(X)= f : X rightarrow mathbbC mid f mathrmis continuous
$$
$$
mathcalC_c(X) = f in mathcalC(X) mid mathrmsuppf mathrm is compact
$$
and if $K subset X$ is compact,
$$
mathcalC(X,K) = f in mathcalC(X) mid mathrmsuppf subset K
$$
Define the topology of $mathcalC_c(X)$ as $varinjlim_K mathcalC(X,K)$.
($varinjlim_K mathcalC(X,K)$ means the colimt in $mathbfTop$ of Banach spaces $mathcalC_c(K)$)
Then, is $varinjlim_K mathcalC(X,K)$ topological vector space?
How to prove?
question.2
Especially when $X= mathbbR^n$, is $varinjlim_K mathcalC(X,K)$ topological vector space?
(I'm sorry my English is broken...I would be glad if you could help me.)
general-topology functional-analysis limits-colimits
question. 1
Let $X$ be a locally compact Hausdorff space, and
$$
mathcalC(X)= f : X rightarrow mathbbC mid f mathrmis continuous
$$
$$
mathcalC_c(X) = f in mathcalC(X) mid mathrmsuppf mathrm is compact
$$
and if $K subset X$ is compact,
$$
mathcalC(X,K) = f in mathcalC(X) mid mathrmsuppf subset K
$$
Define the topology of $mathcalC_c(X)$ as $varinjlim_K mathcalC(X,K)$.
($varinjlim_K mathcalC(X,K)$ means the colimt in $mathbfTop$ of Banach spaces $mathcalC_c(K)$)
Then, is $varinjlim_K mathcalC(X,K)$ topological vector space?
How to prove?
question.2
Especially when $X= mathbbR^n$, is $varinjlim_K mathcalC(X,K)$ topological vector space?
(I'm sorry my English is broken...I would be glad if you could help me.)
general-topology functional-analysis limits-colimits
edited yesterday
asked Aug 4 at 3:41
A. I
637
637
It depends on the category in weich you take the colimit. In the category TVS of topological vector spaces the answer is trivially yes, but in the category of topological spaces it is probably no.
â Jochen
yesterday
Thank you for your reply. Does TVS has the filtered colimit?
â A. I
yesterday
Perhaps math.stackexchange.com/q/1432904 is useful?
â Paul Frost
yesterday
add a comment |Â
It depends on the category in weich you take the colimit. In the category TVS of topological vector spaces the answer is trivially yes, but in the category of topological spaces it is probably no.
â Jochen
yesterday
Thank you for your reply. Does TVS has the filtered colimit?
â A. I
yesterday
Perhaps math.stackexchange.com/q/1432904 is useful?
â Paul Frost
yesterday
It depends on the category in weich you take the colimit. In the category TVS of topological vector spaces the answer is trivially yes, but in the category of topological spaces it is probably no.
â Jochen
yesterday
It depends on the category in weich you take the colimit. In the category TVS of topological vector spaces the answer is trivially yes, but in the category of topological spaces it is probably no.
â Jochen
yesterday
Thank you for your reply. Does TVS has the filtered colimit?
â A. I
yesterday
Thank you for your reply. Does TVS has the filtered colimit?
â A. I
yesterday
Perhaps math.stackexchange.com/q/1432904 is useful?
â Paul Frost
yesterday
Perhaps math.stackexchange.com/q/1432904 is useful?
â Paul Frost
yesterday
add a comment |Â
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It depends on the category in weich you take the colimit. In the category TVS of topological vector spaces the answer is trivially yes, but in the category of topological spaces it is probably no.
â Jochen
yesterday
Thank you for your reply. Does TVS has the filtered colimit?
â A. I
yesterday
Perhaps math.stackexchange.com/q/1432904 is useful?
â Paul Frost
yesterday