All Subspaces of $\mathbb{R}^3$ I'm a student and was recently introduced to Linear Algebra. I was reading a book about it and stumbled upon a sentence:

*

*all of the sub-spaces of $\mathbb{R}^3$ are as following:

*

*All the planes passing through point $(0,0,0)$

*All the lines passing through point $(0,0,0)$

*$\mathbb R^3$ itself

*$\left\{ 0\right\} $
And no proof was given for it, so I tried to come up with a proof. I did manage to prove that these are indeed  sub-spaces of $\mathbb{R}^3$ (sort of), However I don't have any clue how I can prove that these are all  of the sub-spaces (in other words: no other sub-spaces exist). Any help would be greatly appreciated.
Thanks in advance
P.S. sorry for my bad English.
 A: I found a way to prove this problem:
Let's call our subspace $V$.
First of all we know that a subspace can't be an empty set, thus $V\neq\{\}$.
So $V$ has to at least have one member such as $(a,b,c)\in V$, since V is a subspace it should be closed under addition and multiplication, so: $\forall k\in\mathbb{R}:k(a,b,c)\in V$. Since $k$ can also be $0$, we can conclude that every subspace (and space for that matter) must include $0$ : $\forall V\leqslant\mathbb{R^{3}}:(0,0,0)\in V$.
So now we have several cases to investigate:

*

*$|V|=0$:
This case is discussed above.


*$|V|=1$:
We just proved that this concludes $V=\{(0,0,0)\}$, thus proving $\{0\}$ is subspace of $\mathbb{\mathbb{R^{3}}}$. (part 4 in the question)


*$V$ contains at least one more member such as $a$:
Since $V$ is a subspace it should be closed under multiplication, As a result it can easily be shown that all the points on the line connecting the origin point to $a$ (let's call it $l_{1}$) are also in $V$. If $V$ contains no more members than the points on $l_{1}$, then $V$ is a line passing through the origin point. (part 2 in the question)


*$V$ contains at least one more member such as $b$ that is not on $l_{1}$:
Let's call the line connecting $b$ to the origin point $l_{2}$. Similarly to the $3^{rd}$ case it can be shown that all the points on $l_{2}$ are also present in $V$. Let's also call the plane containing $a$, $b$ and the origin point $P_{1}$. Now we claim that all of the points present on $P_{1}$ are also present in $V$. For instance let's say there's a point $p$ on $P_{1}$ that is not on $l_{1}$ nor $l_{2}$. We draw a line from $p$ parallel to $l_{1}$ (called $l'_{1}$), let's say that $l'_{1}$ intersect with $l_{2}$ at point $i_{1}$, similarly we draw $l'_{2}$ and find $i_{2}$. It can be easily shown that the sum of vectors $\overrightarrow{0i_{1}}$ and $\overrightarrow{0i_{2}}$ equals to $\overrightarrow{0p}$, and since $V$ is a subspace (and closed under addition), it also contains $p$. This holds true for any p on plane $P_{1}$, thus $V$ is a plane passing through the origin point. (part 1 in the question)


*$V$ contains at least one more member such as $c$ that is not on $P_{1}$:
let's call the line connecting $c$ to the origin point $l_{3}$. Similarly to the $3^{rd}$ case it can be shown that $V$ contains all the points present on $l_{3}$. We claim that all of the points present in the space (which is $\mathbb{R^{3}}$) are also present in $V$. Let's take an arbitrary point in space such as $q$, and let's call the plane passing through $q$, $c$ and the origin point $P_{2}$, also let's call the intersection of $P_{1}$ and $P_{2}$ $l_{4}$. Now similar to the $4^{th}$ case it can be shown that all the points present on $P_{2}$ are also present in $V$, and since $q$ is an arbitrary point, it can be shown that any point in the space (quick reminder: our space is $\mathbb{R^{3}}$ here), is present in $V$. (part 3 in the question)


*There is at least one more member in $V$ such as $d$ that we haven't considered:
Since at this point $V = \mathbb{R^{3}}$, it means that $d$ is also not present in $\mathbb{R^{3}}$, which results in a contradiction.
Thus we made all the possible $V$s, And in doing so proved the problem.
