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Why is the Euclidean Space $\mathbb{R}^n$ the only possible example of an n-dimensional space?

I am wondering if it is because the existence of the Hamel Base...

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    Who says $\mathbb{R}^n$ is the only possible $n$-dimensional space? – 2017-01-07
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    Please can you comment on that statement? as what it means to say that? – 2017-01-07
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    The question is what _you_ mean by _your_ statement. What you wrote appears to be false, but you may have a very specific context in mind that would change how we should interpret what you wrote. – 2017-01-07
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    Hello, the question is not false. It is 100% true. The only possible example of an n-dimensional space is $\mathbb{R}^n$(or $\mathbb{C}^n$). This can be shown by isomorphism but don't really have a clear idea – 2017-01-07
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    @MichealOguntola It cannot be 100% true if you already have counterexample in one of the answers, the $\mathbb{F}_2^n$ is also $n$ dimensional vector space in a sense. You probably forgot to mention some additional details/assumptions of your problem. – 2017-01-07
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    "Is isomorphic to" does not mean "is". Certainly it is possible to make some very minor changes in wording so that the statement in the question is correct, but this can be done in different ways, and since it's your question you get to choose which of several possible true facts you want to ask about. – 2017-01-07
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    The question http://math.stackexchange.com/questions/842774/how-do-we-know-that-every-n-dimensional-over-bbb-r-is-isomorphic asks about a true fact about vector spaces that sounds much like your question. Is your question the same? – 2017-01-07

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It's not the only space. $\mathbb{C}^n$ is another space with dimension $n$. In fact anytime you have a vector space $\mathbb{F}^n$ over the base field $\mathbb{F}$ you have dimension $n$. Recall that $\mathbb{R}$ and $\mathbb{C}$ form a field. There are, of course, others. Also, every vector space has a basis, and finite-dimensional vector spaces, such as $\mathbb{R}^n$ and $\mathbb{C}^n$) have a Hamel Basis, while infinite-dimensional spaces like $L^p([a,b])$ and $C^{\infty}(\mathbb{R}^n)$ have a Schauder Basis.

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    When we say $\mathbb{R}^n$ in some ways $\mathbb{C}^n$ is covered (by Isomorphism)neglecting the structure of points in them. – 2017-01-07
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    $\mathbb{R}^n$ and $\mathbb{C}^n$ are indeed Euclidean spaces, but that clearly doesn't mean that the they are only spaces of dimension $n$. Also "structure of points" is imprecise. The points (elements) form a field, the difference between the two is when complex conjugation arises, which causes a slightly different algebraic structure. This is probably what you meant to say. – 2017-01-07
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    Either of the two are the only possible example of an n-dimensional space. This has been shown using Isomorphism, I just need a clear idea of it – 2017-01-07
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    This is false, you did not understand anything I wrote. Anytime you have a field, and that doesn't mean either $\mathbb{R}^n$ or $\mathbb{C}^n$, there are literally an infinite other number of fields which have dimension $n$ but are not isomorphic to either of those two spaces. – 2017-01-07
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    Even you don't understand what you are saying. You are telling me that given a scalar field $\mathbb{K}$, then either $\mathbb{R}^n$ or $\mathbb{C}^n$ is not isomorphic to $\mathbb{K}^n$ (in some ways). My friend, that is totally wrong if that is what you mean. Note that both sets are not a singleton – 2017-01-08
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    You asked if there are any $n$-dimensional spaces besides $\mathbb{R}^n$ and $\mathbb{C}^n$. It is true that *all* spaces of finite dimension are isomorphic, but *only* if you consider them over their base field. As an example $\mathbb{R}^n$ has dimension $n$ over the base field $\mathbb{R}$ but this is not true if we consider some other base field like $\mathbb{Q}$. So that only applies when the vector spaces are over the same base field. – 2017-01-08
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    In addition, you asked for answers, because *you* didn't know the answer, not because we didn't. You claimed that the only possible spaces with dimension $n$ are $\mathbb{R}^n$ or $\mathbb{C}^n$. This is false, and you asserted it to be true without proof or even justification, you do not understand how mathematics works. If you make a horribly incorrect claim, at least back it up with some indication you thought it through so we can discuss errors in the reasoning, If you want help, don't pretend you know something when you don't. – 2017-01-08
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    Good point. So, I agree with you that the two spaces are not the ONLY POSSIBLE EXAMPLES of n-dimensional spaces. However, what does it mean to say the two spaces are the only possible examples? Can you give a brief comments on this? – 2017-01-09
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    There is a phrase called "up to isomorphism". So in the context of vector spaces, as an unrelated example, we can say something like "the only two vector spaces that have some properties are $V$ and $W$, up to isomorphism". This means that any other vector space having the properties would be isomorphic to V and W, so it's like saying we don't get any different structures besides V and W. – 2017-01-09
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    Very helpful, now I get the point – 2017-01-10
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All vector spaces of the same finite dimension over the same field are isomorphic. However, there are different spaces of the same dimension given by varying the underlying field (for example, $\mathbb{F}_2^n$).

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    In some ways, the answer to the question can be explained using isomorphism but I don't really have the clear idea about it. – 2017-01-07