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Let $X$ be a vector space over a field $F$, let $$\left\{ \ Y_j \ \colon \ j \in J \ \right\}$$ be a non-empty collection of (vector) subspaces of $X$. Then the intersection $$ \bigcap_{j \in J} Y_j$$ is indeed a subspace of $X$, but the union
$$ \bigcup_{j \in J} Y_j$$ is not necessarily a subspace of $X$.

However, if, for some $j_0 \in J$, we have $$ Y_j \subset Y_{j_0} \text{ for all } j \in J, $$ then of course $$ \bigcup_{j \in J} Y_j = Y_{j_0}$$ and is thus also a subspace of $X$.

Now let's suppose that $$ \bigcup_{j \in J} Y_j$$ is a subspace of $X$. Then can we prove that, for some $j_0 \in J$, $$ Y_j \subset Y_{j_0} \ \mbox{ for all } j \in J?$$

I know that the answer is in the affirmative if the collection consists of only two subspaces.

What is the situation in general?

A rigorous proof will be appreciated.

Michael Hardy
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Saaqib Mahmood
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1 Answers1

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A nice condition for this is that your collection should be "directed". That is, for all $i,j \in J$, there is $k \in J$ such that $$ Y_i \subseteq Y_k\quad\text{and}\quad Y_j \subseteq Y_k . $$ Then it will follow that $$ \bigcup_{j \in J} Y_j $$ is a subspace.

GEdgar
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