In l^?, let Y be the subset of all sequences with only finitely many nonzero ter

Question

In l^?, let Y be the subset of all sequences with only finitely many nonzero terms. Show that Y is a subspace of l^? but not a closed subspace.

Explanation / Answer

As you asked for details : here is the more formal version ! c_00 = { {x_n} | all except finitely many are zero } e.g. 1,0,0,... only the first term is non-zero rest all are zero also 0,0,0 ... is another example. now we are considering it inside l^infty. so our norm is ||x|| = sup {|x_n|}. now take two sequences : x and y from c_00 and any two scalars a,b from R. then I claim ax + by is an element of c_00. How ? : note that when I am multiplying a scalar, it means multiplying the whole seq term by term with the scalar. now x has finitely many non zero terms , collect their positions into a set : { k1,k2,..,kn} same set holds for ax if a is not zero, it has zero in all the rest palces. similarly for y and by we have : {l1,l2,...lm} Now the ax+by has non-zero elements in the positions : [N - {k1,k2,...kn}] U [N- {l1,l2,..lm}] , so the positions of non-zero elements in ax+by is a subset of {k1,...,kn} U {l1,...,lm}, hence finite again. So ax+by is in c_00, 0-sequence belong to c_00, => c_00 is a subspace, in l^infinity. I had given you a wrong example : (oops sorry !! you did the right thing to ask for details ) Correct example : y1 = (1,0,0,...) y2 = (1,1/2,0,0,...) y3 = (1,1/2,1/3,0,0,...) ......... ----> x = (1,1/2, 1/3, 1/4,...) given epsilon > 0 we can always choose n > 0 s.t. 1/n n || x - ym || = sup {|xn - ymn|} < 1/n x in l^infty norm, but x doesnot belong to c_00. Hence it is a subspace which is not closed. Have a good day !!

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