Functional Analysis - subspace of Banach Space

A subspace Y of a Banach space X is a subspace of X considered as a normed space. Hence we don’t require Y to be complete. is this true? I cannot come out with an example to this

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1 Answer

It's true. Take, for instance the set of all sequences in $c$ that are eventually 0 (that is, of the form $(x_1,x_2,\ldots, x_n,0,0,\ldots)$ for some $n$. Here, $c$ is the space of all bounded sequences with limit 0 endowed with the $\sup$ norm.

More generally: It is known that an infinite dimensional Banach space cannot have a countable Hamel Basis. From this, it follows that the set of finite linear combinations of elements from a countably infinite, linearly independent subset of $X$ cannot be closed (closed subspaces of Banach spaces are Banach spaces).

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The fact about Hamel basis, which you mentioned, is shown in one of the answers to this question: math.stackexchange.com/questions/74101/… –  Martin Sleziak Nov 11 '11 at 12:18
yes..i've already edited this twice though... –  David Mitra Nov 11 '11 at 12:35