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The definition I have is the following:

A vector space V is said to be finite-dimensional if there is a finite set of vectors in V that spans V and is said to be infinite-dimensional if no such set exists.

However, with this definition I can't determine whether the vector space $\mathbb{R}^3$ is finite-dimensional or infinite-dimensional (I am assuming that it is finite since the dimension of $\mathbb{R}^3$ is $3$)

Going with my thought process, though, I know that $(1,0,0),(0,1,0),(0,0,1)$ spans $\mathbb{R}^3$. However we can also check that $(2,0,0),(0,2,0),(0,0,2)$ spans $\mathbb{R}^3$. Also note that $(3,0,0),(0,3,0),(0,0,3)$ spans $\mathbb{R}^3$. This process could be continued over and over to show that there are infinitely many vectors that span $\mathbb{R}^3$.

Wouldn't this mean that $\mathbb{R}^3$ is infinite-dimensional? Because there isn't a finite number of vectors that span $\mathbb{R}^3$. (Again I want to say this isn't the case and that there is something I am overlooking.)

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  • $\begingroup$ Your definition should read as "there is at least a finite set of vectors". If there is a finite set, you can generate uncountably many. I'd suggest rephrasing it as "there is a minimal spanning set that is finite". $\endgroup$ – mlc Mar 25 '17 at 22:30
  • $\begingroup$ There are many finite sets of vectors that span $R^3$; the definition only requires at least one exist. $\endgroup$ – Ian Mar 25 '17 at 22:30
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    $\begingroup$ The collection of spanning vectors is not unique. As you noted, the unit vectors $e_1,e_2,e_3$ span $\mathbb{R}^3$ so you are done. It is finite dimensional. There are lots of other collections that span $\mathbb{R}^3$ Including infinite collections), but that is irrelevant. $\endgroup$ – copper.hat Mar 25 '17 at 22:34
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The question asks whether there exists a finite basis of the vector space. If there exists a finite basis, then this vector space is said to be finite dimensional. If not the vector space is infinite dimensional. An example of an infinite dimensional vector space is the vector space of all power series.

Contrast this with the vector space of all polynomials of degree less than or equal to 3, $\mathbb{P}_3 [t]$ which has finite dimension 4 since one basis consists of $\{1,t,t^2,t^3\}$

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  • $\begingroup$ I think I got it now. And since the standard basis for your example is $\{1,x,x^2,x^3\}$ it would be finite-dimensional. $\endgroup$ – WaveX Mar 25 '17 at 22:47
  • $\begingroup$ Precisely! I must have finished my edit after you posted :) $\endgroup$ – cws Mar 25 '17 at 22:49
  • $\begingroup$ Indeed you did ;) $\endgroup$ – WaveX Mar 25 '17 at 22:50
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The dimension of a vector space V is the cardinality of a basis of V over its base field.By a basis we mean a set B of vectors $b_i$, spanning the entire space that is also linearly independent. If this basis is finite, the space is call finite-dimensional. Likewise for an infinite set.

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A vector space is called finite-dimensional, if some list of vectors in it spans the space. By definition, every list has finite length. For every positive integer $n,$ $\mathbb{F^n}$ is a finite-dimensional vector space.

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Definition: A city is said to be food-friendly if one can get three different cuisines in a single restaurant.

In London, there is Restaurant A which serves Indian, Chinese and Continental. ANd also Restaurant B that serves Thai, Mexican and Greek food. And restaurant C serving, Indian, Russian and Japanese.

Now does London meet the above definition of food-friendly city? or do you say there three restaurants A,B, C instead of a single restaurant? So London is not food-friendly??

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