I'm losing my mind over this question. For $H$ a Hilbert space, $A,B$ closed subspaces, and $B$ is of dimension $1$, I want to prove that $A+B$ is also closed.
I'm looking for a straightforward proof, with minimum use of high theorems (without Hahn-Banach, etc).
What I did so far was taking a converging sequence in A+B, and mark it as: $x_k=a_k+\lambda_kv$, for $v$ the base of $B$. I figured out that I could write the limit, $x$, as $x=P_Ax+P_{A^\perp}$ and do the same on $B$, and maybe work something from there.
I also thought that somehow it would be good to take advantage of the sequence $\lambda_k$ which is in $\mathbb{R}$ resp. $\mathbb{C}$. Maybe prove that it converges somehow..
Thanks so much
If $B\subset A$, there is nothing to prove, so assume $B\not\subset A$, and let $b$ such that $B=\mathrm{span}\{b\}$. Of course, $b\not\in A$, and since $A$ is closed, $d(b,A)$ is positive, where $d$ is the metric induced by the norm. Without loss of generality, $d(b,A)=1$, and then for any $t\in\mathbb{R}$ and $a\in A$, $$\|tb+a\|\geq d(tb,A)=|t|.$$
Now let $(x_n)\subset A+B$ be a converging sequence. Write for every $n$ $\;x_n=t_n b+a_n$, where $t_n\in\mathbb{R},a_n\in A$, and it follows from the above paragraph that $(t_n)$ is a converging sequence. It follows then that so is $(a_n)$ and practically we're done.
Note that we didn't use the inner product, so all the above is true for Banach spaces.