Modern and classical definitions of continuity of a function at $x_0$. (James R. Munkres "Analysis on Manifolds") I am reading "Analysis on Manifolds" by James R. Munkres.
Munkres wrote two different definitions of continuity of a function at $x_0$:
Modern(?) definition:

Let $X$ and $Y$ be metric spaces, with metrics $d_X$ and $d_Y$, respectively.
We say that a function $f : X \to Y$ is continuous at the point $x_0$ of $X$ if for each open set $V$ of $Y$ containing $f(x_0)$, there is an open set $U$ containing $x_0$ such that $f(U) \subset V$.

Classical definition:

Continuity may be formulated in a way that involves the metrics specifically. The function $f$ is continuous at $x_0$ if and only if the following holds:
For each $\epsilon > 0$, there is a corresponding $\delta > 0$ such that $$d_Y(f(x), f(x_0)) < \epsilon \text{ whenever } d_X(x, x_0) < \delta.$$
This is the classical "$\epsilon$ - $\delta$ formulation of continuity."

After these definitions, Munkres wrote the following theorem without a proof:

Theorem 3.6(b):
Let $f, g : X \to \mathbb{R}$ be continuous at $x_0$. Then $f + g$ and $f-g$ and $f \cdot g$ are continuous at $x_0$; and $f/g$ is continuous at $x_0$ if $g(x_0) \ne 0$.

My proof for $f + g$ is the following:
Let $V_{f+g}$ be any open set of $\mathbb{R}$ containing $(f+g)(x_0)$.
Then, there exists $\epsilon > 0$ such that if $|y - (f(x_0) + g(x_0))| < \epsilon$, then $y \in V_{f+g}$.
Let $V_f := \{y \in \mathbb{R} | |y - f(x_0)| < \frac{\epsilon}{2}\}$ and $V_g := \{y \in \mathbb{R} | |y - g(x_0)| < \frac{\epsilon}{2}\}$.
Then, $V_f$ and $V_g$ are open sets of $\mathbb{R}$.
So, there exist open sets $U_f$ and $U_g$ of $X$ containing $x_0$ such that $f(U_f) \subset V_f$ and $g(U_g) \subset V_g$.
Let $U_{f+g} := U_f \cap U_g$.
Then if $x \in U_{f+g}$, then $f(x) \in f(U_{f+g}) \subset f(U_f) \subset V_f$ and $g(x) \in g(U_{f+g}) \subset g(U_g) \subset V_g$.
So, $|f(x) - f(x_0)| < \frac{\epsilon}{2}$ and $|g(x) - g(x_0)| < \frac{\epsilon}{2}$.
So, $|f(x) + g(x) - (f(x_0) + g(x_0))| \leq |f(x) - f(x_0)| + |g(x) - g(x_0)| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$.
So, if $x \in U_{f+g}$, then $(f + g)(x) \in V_{f+g}$.
So, $(f+g)(U_{f+g}) \subset V_{f+g}$.
Munkres said:
"Continuity may be formulated in a way that involves the metrics specifically."
and
"This is the classical '$\epsilon$ - $\delta$ formulation of continuity.'"
Since I don't like the word "classical", I didn't want to use metrics specifically in the above proof.
But I had to use metrics specifically.
So my proof is "classical".
Please give me a "modern" proof if it exists.
 A: There are various working descriptions of continuity. Which one you use has not so much to do with "technical", "classical", "modern", etc., but with the mathematical environment you are working in.
If you are working with spaces far away from the real numbers you maybe don't even  have a metric. But you may have a topology, i.e., a system of neighborhoods and open sets that allows to think of "nearness". In such a case you work with the first of Munkre's definition.
As soon as you have a metric $d$ on your space $X$ you can describe the neighborhoods and open sets in terms of the metric, and the continuity can then be described in terms of the auxiliary variables $\epsilon$ and $\delta$.
In analysis we have the real numbers and tuples of them to parametrize the space points. In this case we replace the awkward $d(x,y)$ by the much more intuitive $|x-y|$ or $\|x-y\|$. Continuity is still in terms of $\epsilon$ and $\delta$; but sometimes it is more handy to format neighborhoods  of points $x\in{\mathbb R}$ in the form $\>]u,v[\>$.
As for Theorem 3.6(b) we can set up a more structural proof as follows:

*

*One shows the fundamental principles that the composition of two continuous functions is continuous, and that a function $$f=(f_1,\ldots,f_n):\quad X\to {\mathbb R}^n$$
is continuous iff all component functions $f_i:\>X\to{\mathbb R}$ are continuous.

*One shows that the map $${\rm add}:\quad{\mathbb R}^2\to{\mathbb
   R},\qquad (x,y)\mapsto x+y$$  is continuous. This proof uses the idea of halving you
have used in your suggested proof, but works just with numbers and
the triangle inequality. (Similarly one should do this for all other arithmetic
operations.)

Given these basic facts you have a function $\phi=(f,g):\>X\to{\mathbb R}^2$. When $f$ and $g$ are continuous then $\phi$ is continuous, and therefore $${\rm add}\circ\phi:\quad X\to{\mathbb R},\qquad x\mapsto f(x)+g(x)$$
is continuous.
