Is $\text{span}\{(1, 1, 0, 0), (1, 2, 0, 0)\} = \mathbb{R}^2$? I come with a very simple question, hopefully of sufficient entity not to be disregarded as too elementary for the site. Concretely, I want to know whether
$$\text{span}\{(1, 1, 0, 0), (1, 2, 0, 0)\} = \mathbb{R}^2$$
is a true identity or not. In other words, if a spanning set spans all vectors over $\mathbb{R}^4$ whose third and fourth coordinates are always null, are we correct in stating such space is simply $\mathbb{R}^2$?
On one hand, it seems intuitive to say yes. If we visually depict three-dimensional vectors whose third coordinate is always $0$, we are indeed simply visualizing the cartesian $2$D plane.
On the other hand, $(x_1, x_2, 0, 0)$ is a perfectly valid vector over $\mathbb{R}^4$, fourth-dimensional in full right, and therefore the set of all possible vectors of such form might very well be deemed "properly" four-dimensional.
 A: I'm going to expand on my comment - this is a very, very good question, and I'll admit I had similar ones when I first took linear algebra as a freshman in 2013. (This also makes a good true/false question on a quiz or test!)
So $\mathbb{R}^2$ is really a Cartesian product. Recall that the Cartesian product of two sets $X$ and $Y$, written $X \times Y$ is
$$ X \times Y = \{ (x, y) \  | \ x \in X, \ y \in Y \}$$
that is, the set of ordered pairs. We regard $\mathbb{R}^2$ as the product $\mathbb{R} \times \mathbb{R}$ and similarly for the other Euclidean spaces.
The set $X = \operatorname{span} \{(1, 0, 0, 0), (0, 1, 0, 0) \}$ is not $\mathbb{R}^2$ as each part of the span is an element of $\mathbb{R}^4$ - but it behaves just like our usual two-dimensional space we know and love. In this sense we can say that $X$ is isomorphic to $\mathbb{R}^2$ (and I'll write $X \cong \mathbb{R}^2$ for this).
Recall that an isomorphism is a function $f: X \to Y$ such that $f$ is both one-to-one and onto (bijective) and preserves the operations in the structures $X$ and $Y$. Here, the structure is linearity, so an isomorphism between two vector spaces $V$ and $W$ should obey the rule
$$f(\lambda v_1 + \mu v_2) = \lambda f(v_1) + \mu f(v_2) $$
where $v_1, v_2 \in V$ and $\lambda, \mu \in \mathbb{K}$ (whatever field the vector spaces are over).
To show that our $X$ defined above is isomorphic to $\mathbb{R}^2$, we can choose the candidate function $f: X \to \mathbb{R}^2$ by the rule
$$f \left( (a, b, 0, 0) \right) = (a, b)$$
and I'll leave it as an exercise to show that this map is one-to-one, onto, and obeys the linearity requirement.
As an additional exercise inspired by Ryszard Szwarc's comment - show that any two-dimensional subspace of $\mathbb{R}^4$ is isomorphic to $\mathbb{R}^2.$ (No cheating! Don't just say that the dimensions match!)
Addendum: I'm going to provide a proof of the claim behind a spoiler tag, but in a general sense (i.e., not restricted to subspaces of $\mathbb{R}^4$).
Proposition. Let $V$ be an $n$-dimensional vector space over the field $\mathbb{K}$, and let $W$ be an $m$-dimensional subspace thereof. Then $W \cong \mathbb{K}^m.$
Proof.

 Choose a basis $\{ w_1, \ldots, w_m \}$ of $W$. As this is a linearly independent set that spans $W$, we have a unique representation of any $w \in W.$ That is, there exists a unique set of weights $c_1, \ldots, c_m \in \mathbb{K}$ such that $$w = \sum_{i = 1}^m c_i w_i$$ for any $w \in W.$ Define the function $f: W \to \mathbb{K}^m$ by $$ f(w) = (c_1, \ldots, c_m)$$ which essentially maps $w$ to its coordinates in the standard basis for $\mathbb{K}^m.$ This function is certainly one-to-one - to show this, it's enough to show that $0_W \in W$ has coordinates $(0, \ldots, 0).$ Indeed, this is the case, as $\{ w_1, \ldots, w_m \}$ is linearly independent, so the only solution to $$ \sum_{i = 1}^m c_i w_i = 0_W$$ is when each $c_i = 0.$ To show that this function is onto, notice that any tuple $(c_1, \ldots, c_m) \in \mathbb{K}^m$ can be used to construct a linear combination $$c_1 w_1 + \ldots + c_m + w_m = w \in W.$$ Finally, to show that $f$ preserves linearity, let $x, y \in W$ and $\lambda, \mu \in \mathbb{K}.$ Write $f(x) = (c_1, \ldots, c_m)$ and $f(y) = (d_1, \ldots, d_m).$ Then
 \begin{align} f(\lambda x + \mu y) &= f \left( \lambda \sum_{i = 1}^m c_i w_i + \mu \sum_{i = 1}^m d_i w_i \right) \\ &= f \left( \sum_{i = 1}^m (\lambda c_i + \mu d_i) w_i \right) \\ &= \left( \ldots, \lambda c_i + \mu d_i, \ldots \right) \\ &= \left( \ldots, \lambda c_i, \ldots \right) + \left( \ldots, \mu d_i, \ldots \right) \\ &= \lambda ( \ldots, c_i, \ldots) + \mu ( \ldots, d_i, \ldots) \\ &= \lambda f(x) + \mu f(y) \end{align}
 as required. Thus $f$ is an isomorphism between $W$ and $\mathbb{K}^m.$

