Infimum and absolute values (Rudin's baby analysis) In the second last inequality of the proof of Theorem 6.17 (Rudin's baby analysis), I think he uses the fact that $$|x-y|\leq c \text{ implies } |\inf x-\inf y|\leq c,$$ where $c$ is a real constant and $x,y$ are variables.
I've worked on the proof of this fact for a while, but could not get a clue. Could anyone give me a hint? Currently, I am trying to prove $|\inf x-\inf y|\leq \inf|x-y|$ if this is a correct direction.
 A: I assume you mean that $x$ and $y$ belong to subsets $X$ and $Y$, respectively, of $\mathbb R$.
Let $\epsilon>0$. By definition of $\inf$, there are $x_0\in X;\ y_0\in Y$ such that
$\inf X>x_0-\epsilon$ and $\inf Y>y_0-\epsilon.$ 
And an application of the triangle inequality gives 
$|\inf X-\inf Y|\le |\inf X-x_0|+|x_0-y_0|+|y_0-\inf X|\le 2\epsilon+c.$ 
As $\epsilon$ is arbitrary, the result follows. 
A: It feels like it makes more sense to treat $x$ and $y$ as (real-valued) functions of some parameter $t$; there isn't anything interesting here unless $x$ and $y$ are related to each other somehow.
WLOG, $\inf y\le \inf x$. Now, what can we do with those infimums? Approach one of them - we can find some $t$ such that $x$ is within $\epsilon$ of its infimum, or maybe some $t$ such that $y$ is within $\epsilon$ of its infimum. Which one? Try both, and see which one leads somewhere.
A: I think what you are trying to prove is the following: if $A,B \subset \mathbb R$ and $|a-b| \leq c$ whenever $a\in A$ and $b \in B$ then $|\inf \, A-\inf\, B| \leq c$. To prove this note that $a\leq b+c$ whenever $a\in A$ and $b \in B$. This implies $\inf\, A \leq b+c$ or $\inf\, A -c\leq b$ whenever $b \in B$. In turn, this implies $\inf\, A -c\leq \inf\, B$ or $\inf\, A\leq \inf\, B+c$. Similarly, $\inf\, B\leq \inf\, A+c$. Combining these two we get $|\inf \, A-\inf\, B| \leq c$.
