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