The same matrix can have two different numerical ranks under two declared tolerances

At the tolerance ${\tau }_1=3\mathrm{/}4$, the only singular value strictly above the threshold is ${\sigma }_1=1$, so [L1] gives ${\mathrm{rank}}_{{\tau }_1}(A)=\mathrm{\#}\{j:{\sigma }_j>3\mathrm{/}4\}=1$.

At the tolerance ${\tau }_2=1\mathrm{/}4$, the singular values strictly above the threshold are ${\sigma }_1=1$ and ${\sigma }_2=1\mathrm{/}2$, so [L1] gives ${\mathrm{rank}}_{{\tau }_2}(A)=\mathrm{\#}\{j:{\sigma }_j>1\mathrm{/}4\}=2$.

The singular value ${\sigma }_2=1\mathrm{/}2$ lies strictly between the two tolerances, ${\tau }_2=1\mathrm{/}4<{\sigma }_2<{\tau }_1=3\mathrm{/}4$, which is exactly why it is counted at ${\tau }_2$ but not at ${\tau }_1$.

Steps 1.1, 1.2 and 1.3 exhibit one matrix and two declared tolerances with numerical ranks $1$ and $2$: numerical rank is a function of the matrix together with the tolerance, not of the matrix alone, which refutes the false statement.