Translations, dilations and absorption in a topological vector space

Constant maps are continuous because the preimage of an open set is empty or the whole domain; identity maps are continuous by their preimages. Thus $x\mapsto (a,x)$ and $x\mapsto (x,a)$ into the appropriate products are continuous. Composing with the structure maps proves continuity of translations $x\mapsto x+a$, fixed dilations $x\mapsto bx$, and the orbit maps $s\mapsto sx$ for fixed $x$.

Translation by $-a$ inverts translation by $a$; when $b\mathrm{\ne }0$, dilation by $b^{-1}$ inverts dilation by $b$. The vector axioms and zero/negative identities verify these inverse formulas. The inverses are continuous by step 1.1, so these maps are homeomorphisms. Consequently translates of open sets and nonzero dilates of open sets are open.

Fix an open $U_0$ with $0\in U_0\subseteq U$. Continuity of $s\mapsto sx$ at $0$, where $0x=0$, gives $\delta >0$ such that $\mathrm{\mid }s\mathrm{\mid }<\delta$ implies $sx\in U_0$. For every positive real $t>1\mathrm{/}\delta$, $(1\mathrm{/}t)x\in U$ and hence $x\in tU$. For $x=0$ every positive $t$ works.

Joint addition continuity at $(0,0)$ gives open zero-neighborhoods $U_1,U_2$ with $U_1+U_2\subseteq U_0$. Put $W=U_1\cap U_2\cap (-U_1)\cap (-U_2)$. This is open, contains zero, satisfies $-W=W$, and has $W+W\subseteq U_1+U_2\subseteq U$.

For $\epsilon >0$ put $c=\epsilon \mathrm{/}(M+1)>0$. The set $cN$ is a zero-neighborhood by step 2.1, and $h=cn$ in it satisfies $\mathrm{\mid }f(h)\mathrm{\mid }=c\mathrm{\mid }f(n)\mathrm{\mid }\le cM<\epsilon$. Thus $f$ is continuous at zero. At $x$, the neighborhood $x+cN$ maps into the $\epsilon$-ball about $f(x)$ because $f(x+h)-f(x)=f(h)$. This includes $M=0$ and the zero functional.

For scalar addition at $(a,b)$, errors $\mathrm{\mid }s-a\mathrm{\mid },\mathrm{\mid }t-b\mathrm{\mid }<\epsilon \mathrm{/}2$ give $\mathrm{\mid }(s+t)-(a+b)\mathrm{\mid }<\epsilon$. For multiplication, if $\mathrm{\mid }t-b\mathrm{\mid }<1$ then $\mathrm{\mid }st-ab\mathrm{\mid }\le \mathrm{\mid }s-a\mathrm{\mid }\mathrm{\mid }t\mathrm{\mid }+\mathrm{\mid }a\mathrm{\mid }\mathrm{\mid }t-b\mathrm{\mid }\le (\mathrm{\mid }b\mathrm{\mid }+1)\mathrm{\mid }s-a\mathrm{\mid }+\mathrm{\mid }a\mathrm{\mid }\mathrm{\mid }t-b\mathrm{\mid }\mathrm{.}$ Taking both errors below $\min (1,\epsilon \mathrm{/}(2(\mathrm{\mid }a\mathrm{\mid }+\mathrm{\mid }b\mathrm{\mid }+1)))$ makes this less than $\epsilon$. These are product-open neighborhoods, so they prove joint topological continuity, including $a=0$ or $b=0$. Together with the preceding steps this proves all assertions, without a choice principle or a separation axiom.