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.)







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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.
    – Jochen
    yesterday










  • Thank you for your reply. Does TVS has the filtered colimit?
    – A. I
    yesterday










  • Perhaps math.stackexchange.com/q/1432904 is useful?
    – Paul Frost
    yesterday














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.)







share|cite|improve this question





















  • 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.
    – Jochen
    yesterday










  • Thank you for your reply. Does TVS has the filtered colimit?
    – A. I
    yesterday










  • Perhaps math.stackexchange.com/q/1432904 is useful?
    – Paul Frost
    yesterday












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.)







share|cite|improve this question













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.)









share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








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.
    – Jochen
    yesterday










  • Thank you for your reply. Does TVS has the filtered colimit?
    – A. I
    yesterday










  • Perhaps math.stackexchange.com/q/1432904 is useful?
    – 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.
    – Jochen
    yesterday










  • Thank you for your reply. Does TVS has the filtered colimit?
    – A. I
    yesterday










  • Perhaps math.stackexchange.com/q/1432904 is useful?
    – 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.
– 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.
– Jochen
yesterday












Thank you for your reply. Does TVS has the filtered colimit?
– A. I
yesterday




Thank you for your reply. Does TVS has the filtered colimit?
– A. I
yesterday












Perhaps math.stackexchange.com/q/1432904 is useful?
– Paul Frost
yesterday




Perhaps math.stackexchange.com/q/1432904 is useful?
– Paul Frost
yesterday















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