Proof of The Scaling Set of an Open Convex Neighbourhood of the Origin Has an Infimum

Throughout, balls and openness in $X$ refer to the norm metric $d_{X}$, which is a metric by The Norm Metric of a Real Normed Space (ii), and for $y\in X$ we use $d_{X}(0_{X},y)=\lVert0_{X}-y\rVert=\lVert y\rVert$ (The Norm Metric of a Real Normed Space (i); here $0_{X}-y=-y$ by The Zero Vector, Negatives and Differences in a Vector Space (iii), Vector Space over a Field (ii) and The Zero Vector, Negatives and Differences in a Vector Space (i), and $\lVert -y\rVert=\lVert(-1)y\rVert=\lVert y\rVert$ by Elementary Identities in a Vector Space (iv) and Norms and Real Normed Spaces (iii)).

Since $U$ is open and $0_{X}\in U$, Open Subset of a Metric Space (i) gives a real $r>0$ with $B(0_{X},r)\subseteq U$. Now let $r>0$ be any such radius, let $x\in X$ and let $t>r^{-1}\lVert x\rVert$. Then $t>0$, and by Norms and Real Normed Spaces (iii)

$$d_{X}(0_{X},t^{-1}x)=\lVert t^{-1}x\rVert=t^{-1}\lVert x\rVert<r,$$

so $t^{-1}x\in B(0_{X},r)\subseteq U$ by Open Ball in a Metric Space (ii). Thus $t\in T_{x}$, which proves (i).

Next let $x\in X$, $s\in T_{x}$ and $t>s$; then $t>0$, and we put $\theta=s/t\in(0,1)$. By Vector Space over a Field (v) and Elementary Identities in a Vector Space (iii),

$$t^{-1}x=\theta\,(s^{-1}x)+(1-\theta)\,0_{X},$$

a convex combination of $s^{-1}x\in U$ and $0_{X}\in U$. So $t^{-1}x\in U$ by Convex Subset of a Real Vector Space (i), and $t\in T_{x}$. This proves (ii).

Finally, by (i) the number $r^{-1}\lVert x\rVert+1$ lies in $T_{x}$, so $T_{x}\neq\emptyset$. Every element of $T_{x}$ is positive, so $0$ is a lower bound of $T_{x}$. Hence $T_{x}$ has an infimum by The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness (viii), and $\inf T_{x}\ge0$ because the infimum is the greatest lower bound (Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order (iv)). This proves (iii).