Two terms that I want to understand: weakest topology and jointly continuous (in the following context). I was reading an article online, please help me to understand the following lines (in bold letters). - 
Topological structure:
If (V, ‖·‖) is a normed vector space, the norm ‖·‖ induces a metric and therefore a topology on V.the distance between two vectors u and v is given by ‖u−v‖. 
This topology is precisely the weakest topology which makes ‖·‖ continuous and which is compatible with the linear structure of V in the following sense:
1.The vector addition + : V × V → V is jointly continuous with respect to this topology. This follows directly from the triangle inequality.
2.The scalar multiplication · : K × V → V, where K is the underlying scalar field of V, is jointly continuous. This follows from the triangle inequality and homogeneity of the norm.

Please explain me the Weak topology and how does it makes norm ‖·‖ continuous. what does it mean by "Addition + : V × V → V is jointly continuous with respect to this topology"
Thank you so much in advance. 
 A: The norm $\|\cdot\|$ is a function from $V$ to $\Bbb R$. There are many topologies that can be placed on $V$ that make this function continuous. Let $\mathscr{T}$ be the set of all such topologies. Then it turns out that there is a topology $\tau_0\in\mathscr{T}$ such that $\tau_0\subseteq\tau$ for all $\tau\in\mathscr{T}$. That is, every topology on $V$ that makes the norm a continuous function has to include all of the open sets in $\tau_0$. This, by the way, is equivalent to saying that $\tau_0=\bigcap\mathscr{T}$. Weakest topology here means coarsest topology, i.e., the one with the absolute minimum of open sets needed to make the norm function continuous.
Vector addition is a function $+$ from $V\times V$ to $V$. Once we impose the topology $\tau_0$ on $V$, we automatically get a product topology on $V\times V$, and we can ask whether the function $+:V\times V\to V$ is continuous with respect to that product topology on $V\times V$ and the topology $\tau_0$ on $V$. It turns out that it is. In this context jointly continuous is just a synonym for continuous with respect to the product topology, so (1) is just saying that vector addition is a continuous function. Similarly, the field $K$ has a natural topology $\tau_K$, so $K\times V$ has a product topology defined from $\tau_K$ and $\tau_0$, and (2) is just the claim that the scalar multiplication function
$$\cdot:K\times V\to V:\langle\alpha,v\rangle\mapsto \alpha v$$
is continuous with respect to this product topology in $K\times V$ and the topology $\tau_0$ on $V$.
The reason for the term jointly continuous is that it is possible for a function $f:X\times Y\to Z$ that is not continuous nevertheless to be continuous in each variable separately. That is, it’s possible for each function
$$f_x:Y\to Z:y\mapsto f(x,y)$$
with $x\in X$ to be continuous and for each function
$$f^y:X\to Z:x\mapsto f(x,y)$$
with $y\in Y$ to be continuous, without $f:X\times Y\to Z$ being continuous as a function of two variables.
When each function
$$f_x:Y\to Z:y\mapsto f(x,y)$$
with $x\in X$ to be continuous, we say that $f$ is continuous in the second variable, and when each function
$$f^y:X\to Z:x\mapsto f(x,y)$$
with $y\in Y$ to be continuous, we say that $f$ is continuous in the first variable. If both of these are the case, $f$ is separately continuous, but, as I said, this does not guarantee that it is actually continuous as function from the product space $X\times Y$ to the space $Z$.
A: There is already a very nice answer, but I thought I could give an example of a function which is discontinuous as a function of two variables, but continuous in each variable. Consider the function $f: \mathbb{R}^2 \rightarrow \mathbb{R}$,
$$
f(x,y) = \begin{cases} 
      \frac{xy}{x^2 + y^2} & (x,y) \neq (0,0) \\
      0 & (x,y) = (0,0) \\
   \end{cases}
$$
This is discontinuous as it is 1/2 arbitrarily close to $(0,0)$ when $x=y$, but it is easily seen to be continuous in each variable.
