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So I was reading Linear Algebra: A geometrical approach by S.Kumaresan and there is a problem saying prove that a vector space with finite number of elements in basis will be finite dimensional. Though it is easy to prove, in the very next part they said although in later part you will see that converse is not necessarily true. And here is my problem. Considering what they are saying then a finite dimensional vector space may have infinite number of elements in basis. Consider a vector space $V$ which is finite dimensional(let it be $n$ dimensional) have an infinite set as basis. Now there is a definition that statesA vector space is $k$ dimensional if it has a set of $k$ elements as basis. so the vector space $V$ which is finite dimensional(let it be $n$ dimensional) will have a finite element($n$) set as basis. But it is not true. Since in a finite dimensional vector space's any two bases have same number of elements. hence all of vector space$V$'s basis will have n element which is contradictory by our first assumption. Then we can't say the vector space is finite dimensional. But that seems contradictory too. I mean a vector space can't be finite dimensional and infinite dimensional at the same time because according to definition of finite dimensional: A vector space$V$ is finite dimensional if it has a finite elements set $S$ such that $L(S)=V$. Where $L(S)$ is span of the set $S$. Please tell me where am I wrong in this whole arguement. Thanks in advance.

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  • $\begingroup$ What does "finite number of basis" mean? Does it mean that the size of the basis is finite? $\endgroup$ – астон вілла олоф мэллбэрг Dec 12 '17 at 5:15
  • $\begingroup$ Not infinite number of elements in a basis, but infinite number of finite basis (with all the same number of elements). Take $\mathbb R^2$, there are infinity couple of line who span it. $\endgroup$ – user371663 Dec 12 '17 at 5:15
  • $\begingroup$ @Lucas even $\Bbb R^1$ has infinitely many bases. $\endgroup$ – Lord Shark the Unknown Dec 12 '17 at 5:16
  • $\begingroup$ Yes it is true :) $\endgroup$ – user371663 Dec 12 '17 at 5:17
  • $\begingroup$ Sorry that was finite number of elements in basis instead of finite basis. $\endgroup$ – user426700 Dec 12 '17 at 5:23
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Apparently it seems that I can't put more than two picture in one post.enter image description here

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OK. So these are the pictures of the definitions and theorems I have used on question. definition of finite dimension and the original question. Later part they have said the converse is not true. Which is my question the theorem

enter image description here enter image description here enter image description here

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  • $\begingroup$ These are all pictures $\endgroup$ – user426700 Dec 12 '17 at 7:00
  • $\begingroup$ The text says the converse of exercise 2.3.12 is TRUE. So where’s the confusion? $\endgroup$ – Macavity Dec 12 '17 at 7:39

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