# Concerning Subspaces of Infinite Dimensional Vector Spaces

I have a question concerning subspaces of infinite dimensional vector spaces. Specifically given any infinite dimensional vector space V, how might one construct an infinite decreasing chain of subspaces?

That is:

V=V0$\supseteq$V1$\supseteq$... , where each Vi is properly contained in Vi-1.

I know such chains must exist and I suspect that they should be easily constructed, however I am not familiar enough with infinite dimensional vector spaces to be confident with my attempts thus far.

Thanks for any help given!

## Answers and Replies

Select a basis for the vector space. This should be finite, so we can select a countable chain $\{e_1,e_2,e_3,e_4,...\}$ of basis elements.

Then

$$span\{e_1\}\subseteq span\{e_1,e_2\}\subseteq span \{e_1,e_2,e_3\}\subseteq ...$$

is an infinite chain of subspaces.

Select a basis for the vector space. This should be finite, so we can select a countable chain $\{e_1,e_2,e_3,e_4,...\}$ of basis elements.

Then

$$span\{e_1\}\subseteq span\{e_1,e_2\}\subseteq span \{e_1,e_2,e_3\}\subseteq ...$$

is an infinite chain of subspaces.

Thanks!

So I suppose that with your notation:

$$V = span\{e_1,e_2,...\}\supseteq span\{e_2,e_3,...\}\supseteq...$$ will give me the infinite decreasing chain as desired.

Thanks again for your help!

EDIT: On second thought, can I be assured that such a chain will not stabilize? It is clear in your construction for building increasing chains but it seems less intuitive for building a decreasing chain in such a way.

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If the e_i are components in a basis, then e_n is not in the span of {e_(n+1), e_(n+2),... }, so each inclusion will be proper.