Show that $U\cup W$ is a subspace of $V$ if and only if $U \subset W$ or $W \subset U$

Question

Let $U$ and $W$ be subspaces of $V$. Show that $U\cup W$ is a subspace of $V$ if and only if $U \subset W$ or $W \subset U$.

I am not sure what I can do with the assumption that $U\subset W$ or $W \subset U$ to get started. If I could get a hint pointing me in the right direction, that would be appreciated, thanks

With that assumption, the result immediately follows. If either one of them is inside the other, then the union is just the "biggest" of the two, which we already know is a subSPACE — Francisco José Letterio, Sep 02 '19 at 18:58
In the other direction, if $U\not\subset W$ and $W\not\subset U$, you can choose $u\in U\setminus W$ and $v\in W\setminus U$. Can you then prove from there that $U\cup W$ is not closed under addition? — , Sep 02 '19 at 19:02
@ArturoMagidin that post looks helpful, thanks. I am not sure if it is a duplicate and can let you know after reviewing everything — Evan Kim, Sep 02 '19 at 19:43
@ArturoMagidin yes that post is a duplicate, please close the thread thank you — Evan Kim, Sep 03 '19 at 11:17

score 4 · Answer 1 · answered Sep 02 '19 at 19:21

4

Suppose that $U\cup W$ is a subspace. There are two possibilities

If $U\subseteq W$ then we are done.
If not, let $u\in U\setminus W$ be fixed. We will show that $W\subseteq U$. For every $w\in W$, we have both $u\in U\cup W$ and $w\in U\cup W$, so $u+w\in U\cup W$. Since $W$ is a subspace, $w\in W$ and $u\not\in W$, then $u+w\not\in W$. Therefore $u+w\in U$. Since $u\in U$, then $w=(u+w)-u\in U$ as well.

In both cases, $U\subseteq W$ or $W\subseteq U$.

answered Sep 02 '19 at 19:21

Luiz Cordeiro

18,513

what does the $U \setminus W$ notation mean? – Evan Kim Sep 03 '19 at 11:12
@EvanKim It is set difference: $U\setminus W=\left{u\in U:u\not\in W\right}$. Maybe you've seen it as $U-W$. – Luiz Cordeiro Sep 03 '19 at 12:44

Sundara Narasimhan · Answer 2 · 2019-09-03T16:36:05.050

1

Let us assume that $U\not\subset W$ and $W \not\subset U$ and $U\cup W$ is a subspace.

Then there exist $v \in U$ such that $v\not\in W$ and $w\in W $ such that $w \not\in U$, since $U \cup W $ is a subspace $u+w \in U \cup W$. This implies that $u+w$ belongs to $U$ or in $W$.

But this would force $v \in W$ or $w \in U$. Hence the assertion follows.

edited Sep 03 '19 at 16:36

answered Sep 02 '19 at 19:07

Sundara Narasimhan

640

There is no need to prove this by contradiction; prove it by contrapositive. Your argument shows that if $U\not\subseteq W$ and $W\not\subseteq U$, then your $u+w$ does not lie in $U\cup W$, hence $U\cup W$ is not a subspace. – Arturo Magidin Sep 02 '19 at 19:21
@ArturoMagidin agreed. – Sundara Narasimhan Sep 03 '19 at 16:34
@copper.hat Thank you for pointing out the the error. I have corrected it now. – Sundara Narasimhan Sep 03 '19 at 16:37

score 1 · Answer 3 · answered Sep 03 '19 at 17:09

One direction is straightforward.

Suppose $U \cup W$ is a subspace and $W \not \subset U$. Then there is some $w_0 \in W \setminus U$.

Now pick $u \in U$ and note that $u-w_0 \in U \cup W$ (since a subspace) and $u-w_0 \in W $ (otherwise we would have $w_0 \in U$). Hence $u = (u-w_0) + w_0 \in W$ and so $U \subset W$.

score 0 · Answer 4 · answered Sep 03 '19 at 23:49

Proof $U\setminus W\neq\{\emptyset\}$ and $W\setminus U\neq\{\emptyset\}$ implies $U\cup W$ not a subspace of $V$

Suppose there are $u\in U\setminus W$ and $w\in W\setminus U$. Then $u+w\notin U\cup W$ (we know $u+w\notin U$ because that would imply $w\in U$, being $U$ a subspace, and we can show similarly that $u+w\notin W$). Therefore, there are $u,w\in U\cup W$ such that $u+w\notin U\cup W$, hence $U\cup W$ is not closed under addition and thus not a subspace.

Proof that $U\cup W\le V$ implies $U\subseteq W$ or $W\subseteq U$

This is just another way to prove the same statement as above.

For all $u\in U, w\in W$, we have $u+w\in U\cup W$. For each such $u,w$, either $u+w\in U$ or $u+w\in W$ (or both). If $w\notin U$, then $u+w\notin U$ (because $u+w\in U$ would imply $(u+w)-u=w\in U$), and therefore $u+w\in W$, which implies $u\in W$.

We conclude that the existence of some $w\in W\setminus U$ implies $u\in W$ for all $u$, that is, $U\subseteq W$. Similarly, one proves that if $U\subseteq W$ is false then $W\subseteq U$.

Show that $U\cup W$ is a subspace of $V$ if and only if $U \subset W$ or $W \subset U$

4 Answers4

Proof $U\setminus W\neq\{\emptyset\}$ and $W\setminus U\neq\{\emptyset\}$ implies $U\cup W$ not a subspace of $V$

Proof that $U\cup W\le V$ implies $U\subseteq W$ or $W\subseteq U$