Given arbitrary $a,b$ in the extended reals do we really know that;
If $a-b\le c<\infty$ then $a<\infty$
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Add $b$, that should do it – Nov 05 '18 at 14:10
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@M.Nestor and if b is infinite? – Number4 Nov 05 '18 at 14:11
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1Sure. If $a=\infty$ then, since $\infty - \infty$ is not defined, be would have that $b\neq \infty$ and, by axioms of the extended reals, $a-b=\infty + (-b) = \infty$. – Anguepa Nov 05 '18 at 14:14
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Removing the variable $c$, the problem is equivalent to if $a-b<\infty$ implies $a<\infty$. Note that $a<\infty$ is equivalent to $a\neq\infty$, so for the proof it is sufficient to suppose $a=\infty$ and obtain a contradiction.
Suppose $a-b<\infty$, and consider three cases.
Case 1. If $b\in\mathbb{R}$, then add $b$ to the inequality:
$$a-b<\infty \implies a<\infty+b=\infty$$
Case 2. If $b=-\infty$, the assumption is
$$a+\infty<\infty$$
which has no solution in the extended reals. If by contradiction $a=\infty$ then we have $\infty+\infty<\infty$, or simply $\infty<\infty$, a contradiction.
Case 3. If $b=+\infty$, the assumption is
$$a-\infty <\infty$$
If by contradiction $a=\infty$, then the left side is indeterminate, a contradiction.