If $A$ and $B$ are Banach spaces, then $A+B$ is Banach Let $A$ and $B$ be two Banach spaces continuously contained in a Hausdorff topological space $\mathcal{A}$. Then you can define the spaces $A \cap B$ and $A + B$, the latter being
$$
A + B = \{a+b : a \in A, b \in B\}
$$
with norm
$$
||x||_{A + B} = \inf \{||a||_A + ||b||_B : x = a+b, a \in A, b \in B\}.
$$
I'm trying to show this space is Banach. Here's what I have:
Let $\{x_n\}$ be a sequence in $A+B$ fulfilling $\sum_{n=1}^{\infty} ||x_n||_{A+B} < \infty$. I want to show that $\sum_{n=1}^{\infty} x_n$ converges in $A+B$.
For all $n$, using infimum properties, there are $a_n \in A$, $b_n \in B$ so that $x_n = a_n + b_n$ and:
$$
||x_n||_{A+B} + 2^{-n} \geq ||a_n||_A + ||b_n||_B.
$$
Since $\sum_{n=1}^{\infty} (||x_n||_{A+B} + 2^{-n}) < \infty$, both $\sum_{n=1}^{\infty} ||a_n||_{A} < \infty$ and $\sum_{n=1}^{\infty} ||b_n||_{B} < \infty$, and using that both $A$ and $B$ are Banach:
$$
\sum_{k=1}^{n} a_n \xrightarrow{n\to \infty} a \in A
$$
$$
\sum_{k=1}^{n} b_n \xrightarrow{n\to \infty} b \in B
$$
Taking $x = a+b \in A+B $, we have that
$$
\left\| \sum_{k=1}^{n} x_k - x \right\|_{A+B} = \inf \left\{||a^\prime||_A + ||b^\prime||_B : \sum_{k=1}^{n} x_k - x = a^\prime+b^\prime, a^\prime \in A, b^\prime \in B\right\} \leq \left\| \sum_{k=1}^{n} a_k - a \right\|_{A} + \left\| \sum_{k=1}^{n} b_k - b \right\|_{B}.
$$
And we conclude that $\sum_{k=1}^{n} x_k$ converges in $A + B$ so $A + B$ is complete.
The question I have right now is: is there somewhere in the proof where I should have noted that $A + B$ is contained in a Hausdorff space? I was led to understand that the fact $\mathcal{A}$ is Hausdorff would be important in this proof, but I can't see where it should be used. Thanks in advance for any answers.
 A: For $A+B$ to be defined, $\mathcal A$ had to be a vector space as well. So the correct formulation seems to be $A, B$ are subspaces of a (Hausdorff) topological vector space $\mathcal A$ such that the embeddings $(X,\|\cdot\|_X)\rightarrow\mathcal A$ are continuous for $X=A, B$.
We need to check $\|\cdot\|_{A+B}$ is indeed a norm on $A+B$. In particular, if $\|x\|=0$, there must be $x=0$. To show this, we pick a sequence of decompositions $x=a_n+b_n$ such that $\|a_n\|\rightarrow 0$ and $\|b_n\|\rightarrow 0$, which implies $a_n\rightarrow 0$ in $(A, \|\cdot\|_A)$ and $b_n\rightarrow 0$ in $(A, \|\cdot\|_A)$. By the continuity of the inclusions, $a_n\rightarrow 0$ and $b_n\rightarrow 0$ in $\mathcal A$. By addition is continuous on $\mathcal A\times\mathcal A$, we conclude $x = a_n+b_n\rightarrow 0$. Now, we need the Hausdorff condition to conclude that $x = 0$.
To check homogeneity and subadditivity of $\|\cdot\|_{A+B}$ is straightforward. Given any nonzero scalar $k$ and any decomposition $x=a+b$, we have $\|kx\|_{A+B}\le\|ka\| + \|kb\| = |k| \|a\| + |k| \|b\|$, and taking inf over $x=a+b$, we get $\|k x\|_{A+B} \le |k| \|x\|_{A+B}$, therefore there is also $\|\frac{1}{|k|} kx\|_{A+B} \le \frac{1}{k} \|k x \|_{A+B}$.
Therefore, using your argument, $(A+B, \|\cdot\|_{A+B})$ is indeed a Banach space. The problem now is whether the inclusion of this Banach space to $\mathcal A$ is still continuous. We can use the same idea that if $x_n\rightarrow 0$ according to the norm $\|\cdot\|_{A+B}$, then for sufficiently large $n$, there is decompositions $x_n = a_n + b_n$ such that $\|a_n\|_A, \|b_n\|_B \le \epsilon$, therefore $a_n$ is close to $0$ in $A$, hence in $\mathcal A$, so is $b_n$. So the inclusion is continuous at $0$.
If the inclusions $A, B\hookrightarrow\mathcal A$ are both topological embeddings, is $(A+B, \|\cdot\|_{A+B})\hookrightarrow\mathcal A$ also a topological embedding? The answer is negative as shown by the example in the comment.
However if $A+B$ is complete w.r.t the induced topology from $\mathcal A$, then by the open mapping theorem, the includsion will be an embedding.
