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Given two real-valued functions $f$ and $g$, is it true that $\sup(f-g) \geq \sup(f) - \sup(g)$

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Copied from this answer to a deleted question:

I think this question is asking how to show something like $ \sup_{x\in A}(f(x)+g(x))\le\sup_{x\in A}f(x)+\sup_{x\in A}g(x)\tag{1} $ This is an instance of the fact that the supremum over a set is no smaller than the supremum over a subset. The left hand side of $(1)$ is $ \sup_{\substack{x,y\in A\\x=y}}(f(x)+f(y))\tag{2} $ whereas the right hand side of $(1)$ is $ \sup_{x,y\in A}(f(x)+f(y))\tag{3} $ The set of $x$ and $y$ being considered in $(2)$ is a subset of the $x$ and $y$ being considered in $(3)$, so $(1)$ follows.

Using the result above, we have $ \sup((f-g)+g)\le\sup(f-g)+\sup(g) $ which becomes $ \sup(f-g)\ge\sup(f)-\sup(g) $

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    That answer was deleted when its question was. I have copied over the relevant part.2017-02-13