Suppose $X$ is a vector space and $K$ a convex subset of $X$ with $0\in K^0$. Then $$p_K(x+y) \leq p_K(x)+p_K(y) ,\forall x,y\in X$$ My book offers the following proof:
Suppose $s>p_K(x)$ and $t>p_K(y)$. Then $x/s,y/t\in K$ and because K is convex $$\frac{x+y}{s+t}=\frac{s}{s+t}\frac{x}{s}+\frac{t}{s+t}\frac{y}{t}\in K$$ so $p_K(x+y)\leq s+t$. Here is where I stop understanding : This happens $\mathbf{for\ every\ s>p_K(x),t>p_K(y) \implies p_K(x+y) \leq p_K(x)+p_K(y)}$
I can't get the implication that is written in bold letters. Can someone explain? Thanks in advance.
Now you may take, for any $\varepsilon>0$, $s=p_K(x)+\varepsilon/2$ and $t=p_K(y)+\varepsilon/2$. So you have proven that $$ p_K(x+y)\leq p_K(x)+p_K(y)+\varepsilon. $$ As $\varepsilon$ is arbitrary, you get your inequality.