= Solution
Use the Hausdorff convention for a <locally convex space>: a real vector space $X$ is equipped with a separating family $\mathcal P$ of <seminorms>. Thus each $p$ is nonnegative, subadditive and satisfies $p(tx)=|t|p(x)$, and for every $x\neq0$ some $p\in\mathcal P$ has $p(x)>0$. The topology has a neighborhood basis at $a\in X$ consisting of
$$
a+\{x:p_1(x)<\varepsilon_1,\ldots,p_m(x)<\varepsilon_m\},\qquad p_i\in\mathcal P,\quad\varepsilon_i>0.
$$
Equivalently, it is the coarsest vector-space topology making all these <seminorms> continuous. The separating condition is precisely what makes this topology Hausdorff. If the Hausdorff requirement were omitted, continuous <linear functionals> could not distinguish points in the common kernel of the <seminorms>.
The <continuous dual space> $X^*$ consists of all <continuous linear functionals> $X\to\mathbb R$. We first prove the needed <Hahn-Banach theorem>. Let $p$ be a real-valued <sublinear function> on $X$, meaning $p(a+b)\leq p(a)+p(b)$ and $p(ta)=tp(a)$ for $t\geq0$, and let $g$ be linear on a <vector subspace> $M$, with $g(m)\leq p(m)$. To extend across $v\notin M$, write
$$
\widetilde g(m+tv)=g(m)+tc.
$$
The necessary bounds on $c$ are
$$
L=\sup_{m\in M}\bigl[g(m)-p(m-v)\bigr]\leq c\leq\inf_{n\in M}\bigl[p(n+v)-g(n)\bigr]=U.
$$
They are compatible because
$$
g(m)+g(n)=g(m+n)\leq p(m+n)\leq p(m-v)+p(n+v).
$$
Taking one variable equal to zero also shows that $L,U$ are finite. Choose $c\in[L,U]$. The upper bound proves domination when $t>0$, after dividing $m+tv$ by $t$; the lower bound proves it when $t<0$, after dividing by $-t$. Thus $\widetilde g\leq p$ on $M+\mathbb Rv$. This is the <one-dimensional dominated extension of a real linear functional>.
Order all dominated extensions of $g$ by extension of their domains. A chain has an upper bound obtained by taking the union of the domains and <linear functionals>. <Zorn's lemma> gives a maximal extension, and the one-dimensional construction shows that its domain must be all of $X$. We have therefore proved the real dominated-extension theorem. In particular, when $p$ is a <seminorm>, domination at both $x$ and $-x$ gives \b[$|f(x)|\leq p(x)$].
For $x_0\neq0$, choose a continuous <seminorm> $p$ with $p(x_0)>0$, and define $g(tx_0)=tp(x_0)$ on its one-dimensional span. Then $|g(tx_0)|\leq p(tx_0)$. Extend by the theorem just proved to $f$ with $|f(x)|\leq p(x)$. This bound makes $f$ continuous, and $f(x_0)=p(x_0)>0$. Applying this to the difference of two distinct points proves that \b[$X^*$ separates the points of $X$], the <continuous-dual separation theorem for Hausdorff locally convex spaces>.
For separation from a <closed linear subspace>, choose a basic balanced neighborhood $V$ of zero such that $(x_0+V)\cap Y=\varnothing$. Write
$$
V=\{x:q(x)<1\},\qquad q(x)=\max_{1\leq i\leq m}\frac{p_i(x)}{\varepsilon_i}.
$$
Then $q(x_0-y)\geq1$ for every $y\in Y$. On $Y+\mathbb Rx_0$, define $g(y+tx_0)=t$, which is well-defined since $x_0\notin Y$. For $t\neq0$,
$$
q(y+tx_0)=|t|q(x_0+y/t)\geq|t|=|g(y+tx_0)|;
$$
for $t=0$ the inequality is immediate. The proved extension theorem gives a continuous $f$ dominated by $q$, with
$$
\boxed{f(x_0)=1,\qquad f|_Y=0.}
$$
This proves <separation of a point from a closed linear subspace> without using any unproved Hahn-Banach extension or separation result.
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