The union of three subspaces of a vector space is a subspace iff one of the subspaces is contained in the other I attempted to solve the exercise 1.c.13 of Axler's "Linear Algebra Done Right". The problem is to prove that the union of three subspaces of a vector space $V$ is a subspace of $V$ if and only if one of the subspaces is contained in the other.
I know the question is asked and answered here. According to the comment by the author and the answers in the other page, the statement is false when the field only contains two elements. But my attempts did not explicitly used that assumption and to me it is a strong indication that my proof is flawed, but I could not find where. Can you help me to verify my proof?
In the proof, I used the following lemma which is also another exercise in the book and I could prove it.
Lemma 1 (exercise 12). Suppose $U_{1}$ and $U_{2}$ are subspaces of
$V$. Then $U_{1}\cup U_{2}$ is a subspace of $V$ if and only if
$U_{1}\subseteq U_{2}$ or $U_{2}\subseteq U_{1}$.
I also used a small lemma.
Lemma 2. Suppose $A$, $B$, and $C$ are sets. Then
$A\subseteq C$ and $B\subseteq C$ iff $A\cup B\subseteq C$.
Then the following is the theorem that I attempted to prove.
Theorem (exercise 13). Suppose $U_{1}$, $U_{2}$, and $U_{3}$ are subspaces of $V$. Then
$U_{1}\cup U_{2}\cup U_{3}$ is a subspace of $V$ if and only if
one of $U_{1},U_{2},U_{3}$ contains the other two.
Proof.
($\leftarrow$) I could do this part.
($\to$) Suppose $U_{1}\cup U_{2}\cup U_{3}$ is a subspace of $V$. Since $U_{1}\cup U_{2}\cup U_{3}=\left(U_{1}\cup U_{2}\right)\cup U_{3}=\left(U_{1}\cup U_{3}\right)\cup U_{2}=\left(U_{2}\cup U_{3}\right)\cup U_{1}$, by Lemma 1, we have
$$\begin{align*}
\left(U_{1}\cup U_{2}\subseteq U_{3}\right)\lor\left(U_{3}\subseteq U_{1}\cup U_{2}\right), & \text{ and}\\
\left(U_{1}\cup U_{3}\subseteq U_{2}\right)\lor\left(U_{2}\subseteq U_{1}\cup U_{3}\right), & \text{ and}\\
\left(U_{2}\cup U_{3}\subseteq U_{1}\right)\lor\left(U_{1}\subseteq U_{2}\cup U_{3}\right).
\end{align*}$$
If $U_{1}\cup U_{2}\subseteq U_{3}$ or $U_{1}\cup U_{3}\subseteq U_{2}$
or $U_{2}\cup U_{3}\subseteq U_{1}$, by Lemma 2,
it means one of $U_{1},U_{2},U_{3}$ contains the other two. In the
following, we will consider the case when
$$\begin{align*}
U_{3} & \subseteq U_{1}\cup U_{2},\text{ and}\\
U_{2} & \subseteq U_{1}\cup U_{3},\text{ and}\\
U_{1} & \subseteq U_{2}\cup U_{3}.
\end{align*}$$
Base on these assumptions, we will prove
$$\begin{align*}
U_{1}\cup U_{2} & \subseteq U_{3},\text{ or}\\
U_{1}\cup U_{3} & \subseteq U_{2},\text{ or}\\
U_{2}\cup U_{3} & \subseteq U_{1},
\end{align*}$$
Suppose $U_{2}\cup U_{3}\nsubseteq U_{1}$ and $U_{1}\cup U_{3}\nsubseteq U_{2}$.
Then we have
$$\begin{align*}
U_{2}\nsubseteq U_{1} & \lor U_{3}\nsubseteq U_{1},\text{ and}\\
U_{1}\nsubseteq U_{2} & \lor U_{3}\nsubseteq U_{2}.
\end{align*}$$
Then we will prove $U_{1}\cup U_{2}\subseteq U_{3}$.
We need to prove the four cases but I will only prove the case when $U_{2}\nsubseteq U_{1}$ and $U_{1}\nsubseteq U_{2}$. The other cases can be proved similarly.
Since $U_{2}\nsubseteq U_{1}$ and $U_{1}\nsubseteq U_{2}$, we can choose $a\in U_{2}$ such that $a\notin U_{1}$ and $b\in U_{1}$
such that $b\notin U_{2}$. Since $U_{2}\subseteq U_{1}\cup U_{3}$, $a\in U_{1}\cup U_{3}$.
Since $a\notin U_{1}$, $a\in U_{3}$. Similarly, since $U_{1}\subseteq U_{2}\cup U_{3}$, $b\in U_{2}\cup U_{3}$.
Since $b\notin U_{2}$, $b\in U_{3}$.
Suppose $x\in U_{1}\cup U_{2}$. Then $x\in U_{1}$ or $x\in U_{2}$.

*

*Case $x\in U_{1}$.
Since $U_{1}\subseteq U_{2}\cup U_{3}$,
$x\in U_{2}$ or $x\in U_{3}$. If $x\in U_{3}$, it is what we desired.
We will consider the case when $x\in U_{2}$. Since $x,a,b\in U_{1}\cup U_{2}\cup U_{3}$
and $U_{1}\cup U_{2}\cup U_{3}$ is a subspace of $V$, $x+a+b\in U_{1}\cup U_{2}\cup U_{3}$.

*

*Case $x+a+b\in U_{1}$. Since $x,b\in U_{1}$ and $U_{1}$
is a subspace of $V$, $\left(x+a+b\right)-x-b\in U_{1}$ iff $a\in U_{1}$,
which contradicts $a\notin U_{1}$.


*Case $x+a+b\in U_{2}$. Since $x,a\in U_{2}$ and $U_{2}$
is a subspace of $V$, $b\in U_{2}$, which contradicts $b\notin U_{2}$.


*Case $x+a+b\in U_{3}$. Since $a,b\in U_{3}$ and $U_{3}$
is a subspace of $V$, $x\in U_{3}$.




*Case $x\in U_{2}$. Since $U_{2}\subseteq U_{1}\cup U_{3}$,
$x\in U_{1}$ or $x\in U_{3}$. If $x\in U_{3}$, it is what we desired.
We can prove the case when $x\in U_{1}$ similarly as above.
Therefore, $U_{1}\cup U_{2}\subseteq U_{3}$, as required.
 A: You say:

Suppose $U_{1}\cup U_{2}\cup U_{3}$ is a subspace of $V$. Since $U_{1}\cup U_{2}\cup U_{3}=\left(U_{1}\cup U_{2}\right)\cup U_{3}=\left(U_{1}\cup U_{3}\right)\cup U_{2}=\left(U_{2}\cup U_{3}\right)\cup U_{1}$, by Lemma 1, we have
$$\begin{align*}
\left(U_{1}\cup U_{2}\subseteq U_{3}\right)\lor\left(U_{3}\subseteq U_{1}\cup U_{2}\right), & \text{ and}\\
\left(U_{1}\cup U_{3}\subseteq U_{2}\right)\lor\left(U_{2}\subseteq U_{1}\cup U_{3}\right), & \text{ and}\\
\left(U_{2}\cup U_{3}\subseteq U_{1}\right)\lor\left(U_{1}\subseteq U_{2}\cup U_{3}\right).
\end{align*}$$

Now, Lemma 1 says that if you have two subspaces $V_1$ and $V_2$, and $V_1\cup V_2$ is a subspace, then either $V_1\subseteq V_2$ or $V_2\subseteq V_1$. You seem to be trying to apply Lemma 1 to $U_1\cup U_2\cup U_3$ by, for example, thinking of the union as $(U_1\cup U_2)\cup U_3$, and then thinking of $U_3$ as $V_2$, and thinking of $U_1\cup U_2$ as $V_1$. There's a problem with that:

*

*You don't know that $U_1\cup U_2$ is a subspace.

That means that you cannot yet apply Lemma 1 to the expression $(U_1\cup U_2)\cup U_3$ by setting $V_1=U_1\cup U_2$ and $V_2=U_3$; you would first need to prove that $U_1\cup U_2$ is a subspace. Likewise, you cannot apply it to $(U_1\cup U_3)\cup U_2$, nor to $(U_2\cup U_3)\cup U_1$. So this assertion is unjustified.
Note that, by Lemma 1, the only way that $U_1\cup U_2$ could possibly be a subspace would be if either $U_1\subseteq U_2$ or $U_2\subseteq U_1$. So you would be assuming part of what you are trying to show.
As the rest of your attempt flows from this set of conjunctions/disjunctions, the rest of the argument is doomed and I will not review it. This is probably why your argument does not use anything about the field.
A: If you look at what happens in $V=\Bbb F_2^2$ with the subspaces $U_1=\{(0,0),(1,0)\}$, $U_2=\{(0,0),(1,1)\}$, $U_3=\{(0,0),(0,1)\}$, you'll see that $U_1\cup U_2\cup U_3=V$, but none of your six inclusions in the form $U_i\subseteq U_j\cup U_h$ or $U_i\cup U_j\subseteq U_h$ holds. Essentially, it is due to the fact that none of the $U_i\cup U_j$ (with $i\ne j$) is actually a subspace, and therefore lemma 1 doesn't work.
