How to prove by inudction if $\bigcup_{k=1}^{n}V_{k}$ is a subspace over $\mathbb{C}$ then $\bigcup_{k=1}^{n}V_{k}=V_{j}$ for some j?

Question

Suppose $V_{1},...,V_{n}$ are subspaces of the vector space $W$ (over $\mathbb{C}$). Prove: if $\displaystyle\bigcup_{k=1}^{n}V_{k}$ is a subspace then $\displaystyle\bigcup_{k=1}^{n}V_{k}=V_{j}$ for some $j$.

I know how to prove it when $n=2$, and when $n=3$ it becomes complicated. Is there an easy proof for this?

I read this question here

For subspaces, if $N\subseteq M_1\cup\cdots\cup M_k$, then $N\subseteq M_i$ for some $i$?

But for those answers I can't figure out how they use induction. A comment for Asaf Karagila's answer makes sense: The first statement also holds over finite fields, but how does the induction look like that it works only in characteristic $0$?

Those answers all mentioned use inudction, and my main question is: how to use induction to prove this?

Answer 1

Let $V:=\bigcup_{k=1}^n V_k$, a vector space. Induct on $n$. If $n=1$, $V=V_1$, and we're done. So suppose $n>1$. If $V_n=V$, we're done, so suppose $V_n$ is proper in $V$, and nonempty to avoid trivialities. Pick $x\in V_n$, and $y\in V-V_n$. So $y\in V_k$ for some $k<n$.

Consider the infinite set $S=\{y+cx:c\in \mathbb{C}\}$. (This is where it is important that $\mathbb{C}$ is not a finite field.) Then each $y+cx\not\in V_n$, else $y=(y+cx)-cx\in V_n$. These infinitely many elements are contained among the finite number of subspaces $V_1,\dots,V_{n-1}$, so there must be at least two distinct elements of $S$, say $y+c_1x$ and $y+c_2x$, that are in the same $V_k$ for $k<n$. Taking their difference, $(c_1-c_2)x\in V_k$, so $x\in V_k$ since $c_1-c_2\neq 0$. This shows $V_n\subseteq \bigcup_{k=1}^{n-1}V_k$, hence is redundant. So $V=\bigcup_{k=1}^{n-1}V_k$. Since $V$ is expressed as a union of a fewer number of subspaces, by induction $V=V_j$ for some $j$.