I am trying to see if I understand the idea of a bases as presented in the text by Friedberg, Insel, and Spence. Please correct any and all misconceptions I have in the following and feel free to add insight if necessary:

1.) The definition of the bases of a vector space is a linearly independent set that generates the vector space. The number of vectors in the base are always the same, where when I say "same" I mean that there can be a one to one correspondence between the two bases. So even in an infinite dimensional vector space, the bases will have the "same" number of vectors"

2.) I find the following exercise too trivial for it to be correct: Prove that a vector space is infinite dimensional if and only if it contains an infinite linearly independent set.

If a vector space $V$ is infinite dimensional, then by definition it has a bases of infinite vectors. On the other hand, if a vector space has an infinite linearly independent set S, then $Span(S)\subset V$ but $Span(S)$ has infinite dimension so $V$ must also.

3.) If two bases are given of the "same" size, then the space generated by them will be of equal "size." Specifically, I will be able to find some 1-1 correspondence between the two spaces that are generated.

4.) Heuristically, all we have to do to find the dimension of a finite dimensional vector space is consider the number of different variables needed to express an element of the space.

5.) If $H \subset W$, where $H$ and $W$ are subspaces of $V$ with bases $\alpha$ and $\beta$, then, even in the infinite case, $\alpha \subset \beta$. In particular $(\alpha \cap \beta) \subset \alpha$ and $\alpha \subset (\alpha \cup \beta)$. Does everything here remain true even when we extend to infinite intersections (and unions). Also what are conditions for $\alpha \cap \beta$ to be a basis for $H \cap W$ (same with union)?

  • $\begingroup$ It would be better if you ask one clear question that can be answered. There are too many issues here: first, do you know that every vector space has a base? As far as I know, this only holds if you assume the axiom of choice.. $\endgroup$ – Peter Franek Jun 16 '14 at 19:40
  • $\begingroup$ My book does state that every vector space has a base, although it cannot always be constructed $\endgroup$ – illysial Jun 16 '14 at 19:41
  • $\begingroup$ @illysial Finite dimensional vector spaces (which is the primary focus of linear algebra) always have bases; it is never a problem. If you get into infinite dimensional vector spaces, then they still always have bases, under the assumption of the Axiom of Choice however. Stuff resulting from Choice is generally non-constructive by nature. $\endgroup$ – EuYu Jun 16 '14 at 19:43
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    $\begingroup$ What do you mean when, in (2), you say "it is too trivial for it to be correct?" (2) is correct! $\endgroup$ – user39280 Jun 16 '14 at 19:44
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    $\begingroup$ Ok, the inclusion $\alpha\subseteq\beta$ doesn't hold even for one-dimensional spaces; $H=W=V=\mathbb{R}$, $\alpha=\{1\}$, $\beta=\{2\}$ is a counter-example. $\endgroup$ – Peter Franek Jun 16 '14 at 19:48

Your points 1 and 2 are fine, except that very often a vector space is termed 'infinite dimensional' iff it has no finite basis.

In point 3, simply having a 1-1 correspondence between two spaces is much weaker than being the 'same size' as you say. For example, there is a $1-1$ correspondence between $\mathbb{R}^2$ and $\mathbb{R}$, since they have the same cardinality. But the 'dimension' of $\mathbb{R}^2$ is bigger than the dimension of $\mathbb{R}$, taken as vector spaces over $\mathbb{R}$.

Your point 4 is very loose, and I don't want to comment on the heuristic.

In 5, it's unclear what $\alpha \subset \beta$ means. For example, the space $H$ generated by $(1,0,0)$ is a subspace of the space $W$ generated by $\{(1,1,0), (0,-1,0)\}$, but in terms of set inclusion the former basis is not a subset of the latter basis.

  • $\begingroup$ Would point 3 be better if it was a bijective correspondence? $\endgroup$ – illysial Jun 16 '14 at 19:51
  • $\begingroup$ @illysial: I use bijective correspondence is 1-1 correspondence interchangeably. What difference do you mean? $\endgroup$ – davidlowryduda Jun 16 '14 at 19:52
  • $\begingroup$ I mean to say it would be onto as well $\endgroup$ – illysial Jun 16 '14 at 19:55
  • $\begingroup$ @illysial: Yes, I use bijective correspondence, 1-1 correspondence, being both injective and surjective all to mean the same thing. I still don't quite understand what difference you mean? To be clear, there are bijections of $\mathbb{R}^2$ and $\mathbb{R}$. $\endgroup$ – davidlowryduda Jun 16 '14 at 19:58
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    $\begingroup$ no problem! Good luck! $\endgroup$ – davidlowryduda Jun 16 '14 at 20:05

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