I have tried searching for such spaces but i could only find for spaces whose dimension is less than 2.A vector space V can only have non-trivial subspaces if [tex]\dim V\ge 2[/tex].
This means you can choose two linearly independent vectors u, w, which generate 1-dimensional subspaces U, W respectively. Can [tex]U\cup W[/tex] be a subspace? Hint: try to find a linear combination of u,w that is not in [tex]U\cup W[/tex].
[tex]\dim\mathbb{R}^n=n[/tex], surely you knew that?I have tried searching for such spaces but i could only find for spaces whose dimension is less than 2.
yyat; Do you mean i can obtain a linear subspace V of \mathbb{R}^n such that the union of any subspaces of V is a subspace of V?[tex]\dim\mathbb{R}^n=n[/tex], surely you knew that?
yyat; Do you mean i can obtain a linear subspace V of [tex]\mathbb{R}^n such that the union of any subspaces of V is a subspace of V?[tex]\dim\mathbb{R}^n=n[/tex], surely you knew that?
yyat; Do you mean i can obtain a linear subspace V of [tex]\mathbb{R}^n[/tex], such that the union of any subspaces of V is a subspace of V?[tex]\dim\mathbb{R}^n=n[/tex], surely you knew that?
what about if you are not working with n-dimensional space, can you still find such a space?I think the short answer is No.
thanks so much.If the dimension of the space is less than two then the only subspace are V and {0} as yyat pointed out. Hence your question is answered in this case.
If the dimension of the space is greater or equal to two then consider spaces X and Y generated by linearly independent vectors x and y. x+y does not belong to [tex] X \Cup Y [/tex]. Implying you can't pick any subspaces and the union will be a subspace.