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I'm trying to show that an infimum equals a minimum but am not sure if my proof is valid. If we let A be a non-empty subset of R and define the function $f : \mathbb{R} \; -> \mathbb{R}$ by $f(x) = \inf\{|x-a|:a \in A\}$ then I want to show that for all $x \in \mathbb{R}, \; f(x) = \min\{|x-a|:a \in A\}$.

I started by setting $b=\inf\{|x-a|:a \in A\}$. Then $b\leq d$ for all $d \in \{|x-a|:a \in A\}$ since b is a lower bound. In particular, $b\leq \min\{|x-a|:a \in A\}$ and so $\inf\{|x-a|:a \in A\} \leq \min\{|x-a|:a \in A\}$.

Then I considered that since $b$ is the greatest lower bound, it's bigger than any other lower bound. Now $\min\{|x-a|:a \in A\} \leq d$ for all $d \in \{|x-a|:a \in A\}$ so this is a lower bound. Since $b$ is the greatest lower bound, $b \geq \min\{|x-a|:a \in A\}$

Since we have the inequality going both ways we have that inf=min and so f(x)=min. Is this valid?

Additionally, if I now want to show that f is continuous, can I do the following: Let $c \in \mathbb{R}$ and $\epsilon \gt 0$. Then set $\delta = \epsilon$. Then for all $x \in \mathbb{R}$ with $|x-c|< \delta$ we have $|f(x)-f(c)|=|\min\{|x-a|:a \in A\}-\min\{|c-a|:a \in A\}|$. Now pick the $a \in A$ that minimises these distances and we get: = $||x-a|-|c-a|| \leq |x-a-(c-a)|=|x-c| \lt \delta = \epsilon$.

I'm not sure if I can just choose an a in A like that, or if the rest of the proof is even right. Any assistance would be great!

Also, is it possible to show this function is uniformly continuous?

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  • $\begingroup$ How have you defined "min" and how does it differ from "inf"? In general, the minimum of a set of real numbers does not exist. $\endgroup$ – T. Bongers Oct 3 '17 at 15:03
  • $\begingroup$ Here min is just the smallest number in the set $\endgroup$ – Analysis is fun Oct 3 '17 at 15:05
  • $\begingroup$ The minimum of a set does not usually exist; take $A = (0, 1)$ for example. $\endgroup$ – T. Bongers Oct 3 '17 at 15:07
  • $\begingroup$ Oh that changes things - this was an exercise our lecturer set us, to prove that f(x) = {min{|x-a|: a is an element of A} for all x in the reals. $\endgroup$ – Analysis is fun Oct 3 '17 at 15:12
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Take $A=(0,1) = \{x \in \Bbb R : 0 < x < 1\}$ as mentioned by user296602.

Then, let $f:x \mapsto \inf \{|x-a| : a \in A\}$.

Note that $f(0) = \inf \{|0-a| : a \in A\} = \inf \{a : a \in A\} = \inf A = 0$.

However, $\min A$ doesn't exist.


Let $\varepsilon>0$. Let $\delta=\varepsilon$.

Then, when $|x-x_0|<\delta$:

$$\begin{array}{rcl} |f(x)-f(x_0)| &=& |\inf\{|x-a| : a \in A\} - \inf\{|x_0-a| : a \in A\}| \\ &\le& |\inf\{|x_0-a|+|x-x_0| : a \in A\} - \inf\{|x_0-a| : a \in A\}| \\ &=& ||x-x_0| + \inf\{|x_0-a| : a \in A\} - \inf\{|x_0-a| : a \in A\}| \\ &=& |x-x_0| \\ &<& \varepsilon \\ \end{array}$$

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  • $\begingroup$ Ah I see, thanks! If the min isn't defined how do we show continuity? $\endgroup$ – Analysis is fun Oct 3 '17 at 15:21
  • $\begingroup$ @Analysisisfun here's how it's done half-rigorously (which step needs more justification?) $\endgroup$ – Kenny Lau Oct 3 '17 at 16:04

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