Infimum and absolute values (Rudin's baby analysis)

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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 $|\inf \, A-\inf\, B| \leq c$.