Don't understand proof on how scalar multiplication is continuous with respect to a certain topology?

Question

I am stuck on a proof involving a topological vector space. In particular, a part that shows scalar multiplication is continuous with respect to the topology. I have bolded the part of the proof I'm not sure about.

Let $X$ be a real vector space and let $\mathcal{F}$ be a set of real-valued linear functionals on $X$. Define $$ \mathcal{V}_\mathcal{F} = \{ \bigcap_{i=1}^m f^{-1}((a_i,b_i)) \ | \ m \in \mathbb{N}, f_i \in \mathcal{F}, a_i, b_i \in \mathbb{R}, a_i < b_i, \ \text{for} \ i = 1,\dots, m \}, $$ and note that $\mathcal{V}_\mathcal{F}$ is a basis for the topology $\mathcal{U}_\mathcal{F}$ which is the weakest topology on $X$ such that all $f \in \mathcal{F}$ are continuous.

Proof: Showing scalar multiplication is continuous with respect to $\mathcal{U}_\mathcal{F}$.

Let $V \in \mathcal{V}_\mathcal{F}$, and let $\gamma_0 \in \mathbb{R}$ and let $\tilde{x} \in X$ such that $\gamma_0 \tilde{x} \in V$. Then it follows from the definition of $\mathcal{V}_\mathcal{F}$ that there exists a $\delta > 0$, $\delta \neq |\gamma_0|$, such that $(\gamma_0-\delta)\tilde{x} \in V$, and $(\gamma_0+\delta)\tilde{x} \in V$.

Define $$ U:= \frac{1}{\gamma_0-\delta}V \cap \frac{1}{\gamma_0+\delta}V. $$

Then $U \in \mathcal{V}_\mathcal{F}$, and $\tilde{x} \in U$. Furthermore, if $x \in U$ and $\gamma \in \mathbb{R}$ satisfy $|\gamma - \gamma_0| < \delta$, then $(\gamma_0-\delta)x \in V$ and $(\gamma_0+\delta)x \in V$ and hence $\gamma x \in V$, because $V$ is convex. This proves that scalar multiplication is continuous.

Questions:

1. I don't see how we can say that $V$ is convex? And if it is convex, why does that mean $\gamma x \in V$?
2. How does $\gamma x \in V$ also allow us to conclude that scalar multiplication is continuous? I normally think of continuity in terms of the epsilon-delta concept, so I'm not sure how continuity is being shown in this case?

Answer 1

All sets $V$ of the form $f^{-1}[(a,b)]$ are convex as inverse images of convex sets under linear maps. So all members of $\mathcal{V}_{\mathcal{F}}$ are convex as intersections of convex sets. If $x + c$ and $ x-c$ are both in a convex set $C$, so is $\frac{1}{2}(x-c)+\frac{1}{2}(x+c) = x$

Starting with a point $(\gamma_0, \tilde{x}) \in \mathbb{R} \times X$ and an open basic neighbourhood $V$ of its image $m(\gamma_0,\tilde{x})=\gamma_0\tilde{x}$ under the scalar multiplication map $m$, we found two open neighbourhoods, $U$ for $\tilde{x})$ and an open interval $I$ around $\gamma_0$ such that $ m[I \times U] \subseteq V$.

This shows pointwise continuity in the product topology of $m$ at $(\gamma_0, \tilde{x}))$.