I know how to prove this inequality using bases for the row space, however, my professor presented a proof of
\begin{align*}
\operatorname*{rank}(A+B) \leq \operatorname*{rank}A + \operatorname*{rank}B
\end{align*}
using rank-nullity theorem. He fell sick so he just sent the notes, which I don't understand, can someone please help me.
\begin{align*}
n – \operatorname*{null}(A+B) &\leq 2n – \operatorname*{null}A -\operatorname*{null}B \\
\operatorname*{null}(A+B) &\geq -n + \operatorname*{null}A + \operatorname*{null}B \\
\operatorname*{null}A + \operatorname*{null}B &\leq n + \operatorname*{null}(A+B)
\end{align*}
Then it suffices to prove this inequality.
\begin{align*}
N(A) \cap N(B) \subseteq N(A+B)
\end{align*}
I understand why this is true, but I don't understand the next line
\begin{align*}
\operatorname*{null}A + \operatorname*{null}B \setminus (N(A) \cap N(B)) \leq n
\end{align*}
Not only does this not make sense from the sense of 'where does $n$ come from', it also, if I understand correctly, makes no sense from the mathematical standpoint. He probably meant $\dim(N(A) \cap N(B))$
Hence,
\begin{align*}
\operatorname*{null}A + \operatorname*{null}B \leq n + \dim(N(A) \cap N(B)) \leq n + {\rm null}(A+B)
\end{align*}
This last statement makes sense from the fact above that $N(A) \cap N(B) \subseteq N(A+B)$
If anyone knows how to prove this using rank-nullity theorem or understands what happened here, please let me know.
Best Answer
Assuming $A,B$ are linear maps defined on a vector space $E$ of dimension $n$ (this answers your 'where does $n$ come from', which you should have asked earlier in your post), your professor first used the rank-nullity theorem, to reduce the task to proving $$\operatorname*{null}A + \operatorname*{null}B\le n + \operatorname*{null}(A+B).$$
Then, he implicitely applied Grassmann's formula: for any two subspaces $F,G$ of $E,$ $$\dim(F)+\dim(G)-\dim(F\cap G)=\dim(F+G)\le n$$ to $F=N(A)$ and $G=N(B).$ This is why he (nearly) wrote $$\operatorname{null}(A)+\operatorname{null}(B)-\dim(N(A)\cap N(B))\le n.$$ So you rightly guessed his "$\setminus(N(A)\cap N(B))$" was a typo and meant "$-\dim(N(A)\cap N(B))$".
Since the rest was clear, you are done.