Abuse of notation for infimum and supremum

Question

I would like to take the infimum and supremum of two sets $(\frac{1}{2} e^{8m+4} - 1, e^{8m+4} - 1)$ and $(\frac{1}{2} e^{8m+4}, \frac{3}{2}e^{8m+4})$, but writing

$\sup((\frac{1}{2} e^{8m+4}, \frac{3}{2}e^{8m+4})) < \inf((\frac{1}{2} e^{8m+4} - 1, e^{8m+4} - 1)) \log(\inf((\frac{1}{2} e^{8m+4} - 1, e^{8m+4} - 1))) < p_r$ and $\inf((\frac{1}{2} e^{8m+4}, \frac{3}{2}e^{8m+4}))\log(\inf((\frac{1}{2} e^{8m+4}, \frac{3}{2}e^{8m+4}))) > e^{8m+4}$ for all $m > \frac{1}{4}$.

looks absolutely terrible. What I would like to write is this:

If we let $k \in (\frac{3}{2} e^{8m+4} - 1, 2e^{8m+4} - 1)$ and $r \in (\frac{1}{2} e^{8m+4} - 1, e^{8m+4} - 1)$ then we have $c = k - r + 1 \in (\frac{1}{2} e^{8m+4}, \frac{3}{2}e^{8m+4})$. We observe that $\sup(c) < \inf(r) \log(\inf(r)) < p_r$ and $\inf(c)\log(\inf(c)) > e^{8m+4}$ for all $m > \frac{1}{4}$.

These statements aren't equivalent given that $r$ and $c$ are elements of the respective sets and not the sets themselves. However, I fear that both the first approach and assigning variables to the given sets would make this statement significantly longer than it needs to be.

Would this be considered a meaningful abuse of notation, or should I perhaps rewrite this another way?

Thank you.

Answer 1

This isn't a terrible abuse of notation, but it might take some thought for your reader to figure out what you mean. So I think a better way to do this would be to just give names to the sets themselves, rather than elements of them. If you define $K=(\frac{3}{2} e^{8m+4} - 1, 2e^{8m+4} - 1)$, $R=(\frac{1}{2} e^{8m+4} - 1, e^{8m+4} - 1)$, and $C=(\frac{1}{2} e^{8m+4}, \frac{3}{2}e^{8m+4})$, then you can write with no abuse of notation:

If we let $k \in K$ and $r \in R$ then we have $k - r + 1 \in C$. We observe that $\sup C < (\inf R) \log(\inf R) < p_r$ and $(\inf C) \log(\inf C) > e^{8m+4}$ for all $m > \frac{1}{4}$.

Combined with the definitions of the sets, this is only slightly longer than your proposed version, and it is significantly clearer what it means.

Alternatively, it sounds from context like you might not really care about the sups and infs themselves but, and care only about the relationships between the specific numbers $k$, $r$, and $c$ chosen. So you could avoid talking about sups and infs at all, and instead write something like:

If we let $k \in (\frac{3}{2} e^{8m+4} - 1, 2e^{8m+4} - 1)$ and $r \in (\frac{1}{2} e^{8m+4} - 1, e^{8m+4} - 1)$ then we have $c = k - r + 1 \in (\frac{1}{2} e^{8m+4}, \frac{3}{2}e^{8m+4})$. We observe that $c < r \log(r) < p_r$ and $c\log(c) > e^{8m+4}$ for all $m > \frac{1}{4}$. The first inequality folows from $c<\frac{3}{2}e^{8m+4}$ and $r>\frac{1}{2} e^{8m+4} - 1$, and the second inequality follows from $c>\frac{1}{2} e^{8m+4}$.