Is $\mathbb{R}^1$ a subspace of $\mathbb{R}^2?$ My intuition tells me it is. 
But in terms of vectors, the span of a vector with only one component (a vector in $\mathbb{R}^1$) is not said to be a subspace of $\mathbb{R}^2$
 A: Technically, the answer is no, because $\mathbb{R}^1$ is not a subset of $\mathbb{R}^2$. $\mathbb{R}^1$ consists of real numbers, whereas $\mathbb{R}^2$ consists of ordered pairs of real numbers; therefore $\mathbb{R}^1$ is not contained in $\mathbb{R}^2$.
However, there are many ways one can "put a copy of $\mathbb{R}^1$ into $\mathbb{R}^2$", and depending on how this is done, the result may or may not be a subspace of $\mathbb{R}^2$. Look at the definition of a subspace, and consider that 
$$S=\{(x,0)\in\mathbb{R}^2\mid x\in\mathbb{R}^1\}$$
is a subspace of $\mathbb{R}^2$ because  


*

*the zero vector, $(0,0)\in S$   

*for any two $(a,0),(b,0)\in S$, we have that $(a,0)+(b,0)=(a+b,0)\in S$

*for any $(a,0)\in S$ and $c\in\mathbb{R}$, we have that $c(a,0)=(ca,0)\in S$


while
$$T=\{(x,1)\in\mathbb{R}^2\mid x\in\mathbb{R}^1\}$$
completely fails to be a subspace of $\mathbb{R}^2$ - every condition is false:


*

*the zero vector, $(0,0)\notin T$   

*for any two $(a,1),(b,1)\in T$, we don't have that $(a,1)+(b,1)=(a+b,2)\in T$

*for any $(a,1)\in T$ and $c\in\mathbb{R}$, we don't have that $c(a,1)=(ca,c)\in T$ (unless $c=1$)


Generally, you should note that it is entirely possible for a subspace of a vector space $V$ to have smaller dimension than $V$; for example, for any vector space $V$, the subset consisting of only the zero element is always a subspace, and it has dimension 0. 
