Every subspace contains zero, so their intersection does too. If $x,y$ lie in every $W_i$, each combination $ax+by$ lies in every $W_i$, proving closure. The argument works for arbitrarily many subspaces. An empty intersection is conventionally the whole ambient space.
Parents
Like how this is organised?
Build your own knowledge tree on knowen.md: notes, decisions and tasks in one connected markdown graph, shared with your team and your AI agents.