I don’t like how the pairs $(h, h^‘)$ are implicitly the same as $(h^’, h)$. I think the post confuses the pairs with small sets. Normally, order matters in a pair.
The formatting on your comment seems have gotten a little messed up, so appologies if I’ve missed your point, but the paper seems quite specific that the pair (h,h') is not equivalent to (h',h), especially in term of Def 1. For instance, the result of f(h,h') may be positive while the result of f(h',h) is negative or zero.
The formatting you’re seeing is controlled by carets (^). Just replace the start and end of the superscript text with them.
I see this often when talking about exponents: x^2 is less than x^3.
The superscript needs to be surrounded to work properly: x(x+1) is greater than x(x-1).
In this case, the effected text is $(h, h^‘)$ and $(h^’, h)$. I’m not sure why the prime notation needs to be superscripted, probably something to do with the formatter that is expectinng $s. I think the correct formatting (at least for me) is (h, h’) and (h’, h).
I don’t like how the pairs $(h, h^‘)$ are implicitly the same as $(h^’, h)$. I think the post confuses the pairs with small sets. Normally, order matters in a pair.
The formatting on your comment seems have gotten a little messed up, so appologies if I’ve missed your point, but the paper seems quite specific that the pair
(h,h')is not equivalent to(h',h), especially in term of Def 1. For instance, the result off(h,h')may be positive while the result off(h',h)is negative or zero.The formatting you’re seeing is controlled by carets (^). Just replace the start and end of the superscript text with them.
I see this often when talking about exponents: x^2 is less than x^3.
The superscript needs to be surrounded to work properly: x(x+1) is greater than x(x-1).
In this case, the effected text is $(h, h^‘)$ and $(h^’, h)$. I’m not sure why the prime notation needs to be superscripted, probably something to do with the formatter that is expectinng $s. I think the correct formatting (at least for me) is (h, h’) and (h’, h).