Arbitrary products of the scalar field are locally convex

The function $i\mapsto 0$ is a specified element of ${\mathbb{K}}^I$, and $(x+y)(i)=x(i)+y(i)$, $(ax)(i)=ax(i)$ and $(-x)(i)=-x(i)$ define functions on $I$. Associativity, commutativity and the zero/inverse laws follow at each coordinate from the scalar field; the two distributive laws, associativity of scalar action and the unit action also follow at each coordinate. Function equality is coordinatewise, so these give every vector-space axiom.

The $i$-th component of addition is $(x,y)\mapsto x_i+y_i$, a composite of the continuous coordinate maps and scalar addition. The $i$-th component of scalar multiplication is $(a,x)\mapsto a{x}_i$, similarly continuous by joint scalar multiplication. Therefore both vector operations are jointly continuous by product universality. This uses neither projection surjectivity nor nonemptiness of an arbitrary product of unrelated factors.

Each $N(F,\epsilon )$ is open, being a finite intersection of inverse images of open scalar disks, and contains zero. If $x,y$ are in it and $0\le t\le 1$, then $\mathrm{\mid }(1-t)x_i+t{y}_i\mathrm{\mid }\le (1-t)\mathrm{\mid }{x}_i\mathrm{\mid }+t\mathrm{\mid }{y}_i\mathrm{\mid }<{\epsilon }_i$ for $0<t<1$, with $t=0,1$ immediate. If $\mathrm{\mid }a\mathrm{\mid }\le 1$, then $\mathrm{\mid }a{x}_i\mathrm{\mid }\le \mathrm{\mid }{x}_i\mathrm{\mid }<{\epsilon }_i$, including $a=0$. Thus these neighborhoods are convex and balanced. Given any basic zero-neighborhood, choose a positive radius inside each of its finitely many coordinate neighborhoods, by finite choice after listing those coordinates. The resulting $N(F,\epsilon )$ is contained in it. Hence these sets form a base and the TVS is locally convex. For $F=\mathrm{\varnothing }$, the set is the whole space.

If $x\mathrm{\ne }y$, some coordinate $i$ has $d=\mathrm{\mid }{x}_i-y_i\mathrm{\mid }>0$. The inverse images of the disks of radius $d\mathrm{/}3$ about $x_i,y_i$ are open neighborhoods of $x,y$. They are disjoint, since a common scalar value would imply $d<2d\mathrm{/}3$ by the triangle inequality. Thus the space is Hausdorff. For $I=\mathrm{\varnothing }$, its only element is the empty function; its only zero-neighborhood is the whole singleton, the vector operations are constant, and the Hausdorff assertion is vacuous.