$f: X \times Y \mapsto \mathbb{R}$ and $g: Y \mapsto \mathbb{R}$ are continuous, then $f+g$ is continuous on $X \times Y$ I am trying to prove the following statement:
Let $\mathcal{X}$ and $\mathcal{Y}$ be spaces in $\mathbb{R}^d$, suppose $f: \mathcal{X} \times \mathcal{Y} \mapsto \mathbb{R}$ and $g: \mathcal{Y} \mapsto \mathbb{R}$ are continuous, then $f+g: \mathcal{X} \times \mathcal{Y} \mapsto \mathbb{R}$ is continuous.
Attempted proof:
Fix $x\in \mathcal{X}$, $f(x,y)$ and $g(y)$ are continuous with respect to $y \in \mathcal{Y}$. Therefore, $f(x, y) + g(y)$ is continuous with respect to $y \in \mathcal{Y}$.
Likewise, fix $y \in \mathcal{Y}$, $f(x, y)$ is continuous with respect to $x$ and $g(y)$ is a consant. Trivially, $f(x, y) + g(y)$ is continuous with respect to $x \in \mathcal{X}$.
In other words, since $f(x, y)$ and $g(y)$ are continuous, then fix $x \in \mathcal{X}$, $f+g: \{x\} \times \mathcal{Y} \mapsto \mathbb{R}$ is continuous. Likewise, fix $y \in \mathcal{Y}$, $f+g: \mathcal{X} \times \{y\} \mapsto \mathbb{R}$ is continuous.
By definition, fix $x \in \mathcal{X}$, $\forall~ y \in \mathcal{Y}$, $\forall~\epsilon>0$, there exists $\delta>0$ such that $\|(x,y) - (x, y')\|=\|(0, y-y')\|=\|y - y'\| < \frac{\delta}{2}$ implies $|f(x, y) + g(y) - f(x, y') - g(y')| < \frac{\epsilon}{2}$.
Fix $y' \in \mathcal{Y}$ (Same $y'$ when $x$ is fixed), $\forall~x \in \mathcal{X}$, $\forall~\epsilon>0$, there exists $\delta>0$ such that $\|(x, y')-(x',y')\| = \|(x-x',0)\|=\|x - x'\|< \frac{\delta}{2}$ implies $|f(x, y') + g(y') - f(x',y') - g(y')|<\frac{\epsilon}{2}$.
Observe the two inequalities below:
\begin{align}
\epsilon &> |f(x, y) + g(y) - f(x, y') - g(y')| + |f(x, y') + g(y') - f(x',y') - g(y')|\\
 &\geq |f(x, y) + g(y) - f(x, y') - g(y') + f(x, y') + g(y') - f(x',y') - g(y')|\\
&=|f(x,y)+g(y) - f(x',y')-g(y')|
\end{align}
and
\begin{align}
\delta &> \|(0, y-y')\|+\|(x-x', 0)\|\\
&\geq \|(x-x', y-y')\|\\
&=\|(x,y)- (x',y')\|
\end{align}
Therefore, for all $(x, y) \in \mathcal{X} \times \mathcal{Y}$, $\forall ~\epsilon > 0$, there exists $\delta > 0$ such that $\|(x, y) - (x', y')\| < \delta$ implies $|f(x, y) + g(y) - f(x',y') - g(y')|< \epsilon$. By definition, $f+g: \mathcal{X} \times \mathcal{Y} \mapsto \mathbb{R}$ is continuous.       $\qquad\blacksquare$
Sorry for the lack of conciseness in my proof. I tried to make my proof as clear as possible. There are some details which I am not too sure about. e.g:
(1) I have not specified the norm I use, I think $\|(0, y-y')\| = \|y - y'\|$ holds for all norms (at least for p-norm). Is it true? If not, how should I specify the norm so that this holds?
(2) When fixing $x \in \mathcal{X}$, given $f(x,y)$ and $g(y)$ are continuous with respect to $y$, can I claim that $f+g$ is continuous on $\{x\} \times \mathcal{Y}$?
Please let me know if the proof makes sense. If not, how should I properly prove it? Thank you for your time and help. I will appreciate it.
 A: It's already great to be familiar with $\delta$-$\varepsilon$ proofs, but let me give an alternative type of proof, which is a bit more topological, and skips (almost) all the analysis.
Edit: I had written a proof which contained a mistake, as rightfully pointed by Brevan, so let me provide a revised version:
You have two continuous maps: $f: X\times Y \to \mathbb{R}$, and $g : Y \to \mathbb{R}$.
For simplicity, consider the map $\hat{g} : X \times Y \to \mathbb{R} : (x,y) \mapsto (0, g(y))$ (so you pretty much extend $g$ to $X\times Y$ in a natural way). This ensures that $f + g = f + \hat{g}$, but now you're working with two maps from the same domain.
Now, $f$ and $\hat{g}$ are continuous, so it suffices to prove that the sum of two continuous maps $X \times Y \to \mathbb{R}$ is continuous.
On $\mathbb{R}\times\mathbb{R}$, you can define the addition map:
\begin{equation}
+ : \mathbb{R}\times\mathbb{R} \to \mathbb{R} : (x,y) \mapsto x + y
\end{equation}
You can show that this map is continuous. So then, you can express $f + \hat{g}$ as the following composition:

Where $i_1$ is the inclusion of $\mathbb{R}$ into the first coordinate of $\mathbb{R}^2$, and similarly for $i_2$. The composition of continuous maps is continuous, so $\hat{g}$ is continuous.
A: Okay, let me address both of your questions before I give some information of my own.

*

*No, the norm $\left|\left|(0,y-y')\right|\right|$ is not always going to be the same as $\left|\left|y-y'\right|\right|$. However, you are working exclusively with finite-dimensional normed spaces and all norms are equivalent on these spaces. In other words, even if the norm on $X \times Y$ does not reduce to the norm on $Y$ when the first component is $0$, you can always just replace it with an equivalent norm.


*If you fix $x \in X$, then you can just define a function:
$$\phi: Y \to \mathbb{R} \ , \ y \mapsto \phi(y) := f(x,y)$$
Then, you want to show that $\phi+g$ is continuous. You can't claim that $g$ is continuous on $\{x\} \times Y$ because $g$ is not defined on this space (see some of the comments above). On the other hand, there is a sense in which you can define $g$ on this space. Basically, you define a function:
$$\psi: X \times Y \to \mathbb{R} \ , \ (x,y) \mapsto \psi(x,y) := g(y)$$
Once you've done this, then you can talk about adding $\psi$ and $f$, for example, without any issues.
Okay, with all that being said, let me actually try to address your questions now. So, I'm also not really sure why you're working on the product spaces. The product space will be a normed vector space and so, if $V$ is a real normed space, then it is sufficient to show that when $f,g: V \to \mathbb{R}$ are continuous, then $f+g$ is continuous. Let $x_0 \in V$ be fixed. Then:
$$|(f(x)+g(x))-(f(x_0)+g(x_0))| \leq |f(x)-f(x_0)| + |g(x)-g(x_0)|$$
As $x \to x_0$, the right-hand side goes to $0$ because $f$ is continuous at $x_0$ and $g$ is continuous at $x_0$. So, we're done.
