Let $U_1, U_2$ be subspaces of the real vector space $V$. Prove that $U_1 \cup U_2$ is a subspace of $V$ iff $U_1 \subset U_2$ or $U_2 \subset U_1$

Question

Let $U_1, U_2$ be subspaces of the real vector space $V$. Prove that $U_1 \cup U_2$ is a subspace of $V$ iff $U_1 \subset U_2$ or $U_2 \subset U_1$

Is my proof valid?

1. Firstly proving $U_1 \cup U_2 \space \text{is a subspace} \implies U_1 \subset U_2$ or $U_2 \subset U_1$

$\begin{align*}U_1 \cup U_2 \space \text{is a subspace} & \implies \alpha u_1+\beta u_2 \in U_1 \cup U_2 \space\space\forall u_1 \in U_1, \forall u_2 \in U_2, \forall\alpha,\beta \in \mathbb{R} \\ & \implies \alpha_0 u_1+\beta_0 u_2 \in U_1 \space\text{or} \space \alpha_0 u_1+\beta_0 u_2 \in U_2 \space \text{for some $\alpha_0, \beta_0 \neq 0$} \end{align*}$

$\text{WLOG assume}\space \alpha_0 u_1+\beta_0 u_2 \in U_1$

$\begin{align*} & \alpha_0 u_1+\beta_0 u_2 &&\in U_1 \\ \implies & \alpha_0 u_1+\beta_0 u_2 + (-\alpha_0)u_1 &&\in U_1 \\ \implies & \beta_0 u_2 &&\in U_1 \\ \implies & \frac{1}{\beta_0}\cdot\beta_0 u_2 &&\in U_1 \\ \implies & u_2 &&\in U_1 \text{(All by the properties defined on a vector subspace.)} \end{align*}$

Hence $U_2 \subset U_1 \space \text{and similarly} \space U_1 \subset U_2.$ $\text{Thus} \space U_1 \cup U_2 \space \text{is a subspace} \implies U_1 \subset U_2$ or $U_2 \subset U_1$

2. Secondly proving $ U_1 \subset U_2$ or $U_2 \subset U_1 \implies U_1 \cup U_2 \space \text{is a subspace}$

$\text{WLOG assume $U_1 \subset U_2$}$

$\begin{align*} \text{Hence} \\ &u_1 &&\in U_2 \space \forall u_1 \in U_1 \\ &\alpha u_1+\beta u_2 &&\in U_2 \space \forall u_2 \in U_2, \space \forall \alpha, \beta \in \mathbb{R} \\ &\alpha u_1+\beta u_2 &&\in U_1 \cup U_2 \end{align*}$

Hence $U_1 \subset U_2$ or $U_2 \subset U_1 \implies U_1 \cup U_2 \space \text{is a subspace}$

Please let me know if this is correct.

Answer 1

Your proof of the first part is far from clear (see the discussion in the comments) and there is a quicker way to do the second part:

1. Assume that $U_1 \cup U_2$ is a subspace of $V$ and asssume for a contradiction that $U_1 \not\subset U_2$ and $U_2 \not\subset U_1$. Then we can find $u_1 \in U_1 \setminus U_2$ and $u_2 \in U_2 \setminus U_1$. But then $u_1$ and $u_2$ are both members of $U_1 \cup U_2$ and hence $u_1 + u_2 \in U_1 \cup U_2$, since we are assuming $U_1 \cup U_2$ is a subspace of $V$. But then we must have either $u_1 + u_2 \in U_1$ or $u_1 + u_2 \in U_2$. In the first case, we have $u_2 = (u_1 + u_2) - u_1 \in U_1$ contradicting our assumption that $u_2 \in U_2 \setminus U_1$. Similarly we get a contradiction if $u_1 + u_2 \in U_2$. So our assumptions cannot hold: if $U_1 \cup U_2$ is a subset, then either $U_1 \subset U_2$ or $U_2 \subset U_1$.

2. WLOG, we can assume that $U_1 \subset U_2$, but then $U_1 \cup U_2 = U_2$, which is given to be a subspace.

Answer 2

The first part of your proof is incorrect. You cant assume that the sum is in $U_1$, if you were only talking about one object $u_1$ you could assume WLOG that it is in $U_1$, but you are speaking of lots of objects, it may be the case that some of them are in $U_1$ and some of them are in $U_2$. Try contradiction instead.

Assume for a contradiction that $U_1\cup U_2$ is a subspace while $U_1\not\subset U_2$ and $U_2\not\subset U_1$.

Choose any $x$ such that $x\in U_1$ while $x\not\in U_2$. Chose any $y$ such that $y\not\in U_1$ while $y\in U_2$. The assumption that $U_1\cup U_2$ is a subspace implies that $x+y\in U_1\cup U_2$, meaning that $x+y\in U_1$ or $x+y\in U_2$, which is clearly a contradiction.

As for the second part of the proof, assumign WLOG $U_1\subset U_2$ , this implies $U_1\cup U_2= U_2$

Answer 3

The proof seems more or less correct, but more importantly it is very difficult to read and follow, I won't attempt to edit it as that might get too messy, so I'll rewrite your proof here with some improvements and comments for you to learn from:

Let $U,V$ be subspaces of a vector space $X$, then assume that $V\cup U$ is a subspace of $X$, we want to conclude that either $V\subseteq U$ or $U\subseteq V$

Note that the reader is being guided through the thought process, rather than just being confronted with a heap of symbols

To show that either $V\subset U$ or $U\subseteq V$, we start with an arbitrary $u \in U$ and an arbitrary $v\in V$

This is somewhat nonstandard, but actually quite a nice way of doing it, by choosing an arbitrary element from both, we postpone the decision of which inclusion we're going to try and prove

Then from the definition of a subspace we have that

$$v+u \in V\cup U$$

Notice that I got rid of the $\alpha$'s and $\beta$'s, afterall you also get rid of them, just through a much more complicated procedure

Then from the definition of union, either

$$v+u\in V \ \mbox{ or } \ v+u\in U$$

Assume w.l.o.g that

$$ v+u \in V$$

Then since $v\in V$, we also have that $-v\in V$, so we have that

$$(v+u)+(-v) = u\in V$$

So given an arbitrary pair of elements $u,v$ in $U$ and $V$ respectively, either $u\in V$ or $v\in U$. Now If $U\not\subseteq V$ then there is some $u_0 \in U$ such that $u_0 \notin V$, then pairing each $v\in V$ with $u_0$ we find that $V\subseteq U$, we could also conclude that

$$V\not\subseteq U\implies U\subseteq V$$

So the proof is complete.

The idea you had is actually quite elegant when cleaned up a bit, so well done. For the other direction of implication however, there is a significantly shorter proof, notice that

$$A\subseteq B \implies A\cup B = B$$

For arbitrary sets. I'll let you figure out the rest