Prove that $\inf G \in G$ I have the following question
Suppose that G is a nonempty finite set of $\mathbb{R}$ Prove that $\inf G \in G$
My current though is is that
if we use the induction of n, the number of elements of $G$. If $G$ has some element that is $G = \{x\}$ Then $\inf G = x \in G$.
 A: This solution assumes that you are allowed to use the following fact:
For any non-empty finite subset $S$ of $\mathbb{R}$, the minimum of $S$ exists.
For a proof of this fact, see this old question:
A finite set always has a maximum and a minimum.
Hint: Since $G$ is nonempty and finite, its minimum $\min G$ exists. Because $\min G\in G$ by definition, all you need to show is that $\min G =\inf G$.

Now you can follow the definition of the infimum. Let $a=\min G$.

*

*Then for any $g\in G$, you have $a\le g$. Hence $a$ is a lower bound of $G$.


*Now suppose $a'$ is another lower bound of $G$. Since $a\in G$, you have $a'\leq a$.
To conclude that proof, note that the above argument shows that $a$ is the greatest lower bound for $G$.
A: A property of the real numbers is that of connexity:
For all pairs of real numbers (a, b), either (a <= b) or (a >= b).
Thus, for any finite set F, all possible pairs of numbers can get compared such that there exists a number n such that n <= x, for all x in F.
Since, that holds there exists a number n which is the minimum of F.
From here show that any minimum of a set is also an infimum of that set.
A: This is true for all sets with order $k=1$.
If $|G| = 1$ then there is an $x\in \mathbb R$ so that $G= \{x\}$.  Now $x \le x$ as $x$ is the only element in $G$ that means $x \le a$ for all $a\in G$.
If we assume it is true for all sets of order $k$:
Let $|G| = k+1$.  Let $x_0\in G$.  Then $|G\setminus\{x_0\}| = k$ so $G\setminus\{x_0\}$ has a minimum element.  Let's label it $m$.
$x_0\ne m$ as $x_0 \not \in G\setminus\{x_0\}$ and the elements of a set must be distinct. So either $x_0 > m$ or $x_0 < m$.
If $x_0 > m$ then for any $a \in G= G\setminus\{x_0\}\cup \{x_0\}$ either $a = x_0$ and $m < a_0$.  Of $a\ne x_0$ and $a \in G\setminus\{x_0\}$ and $m = \min G\setminus\{x_0\} \le a$.  Either way $m \le a$ so $m = \min G$.
If however $x_0 < m$ then for any $a\in G$ either $a=x_0$ and $x_0 \le a$ or $a\in G\setminus\{x_0\}$ and $x_0 < m \le a$.  Either way $x_0 \le a$ so $x_0 = \min G$.
So $\min G$ always exist.
And as $\min G \le a$ for any $a \in G$ then $\min G$ is a lower bound of $G$.  And if $k> \min G$ then $\min G \in G$ and $\min G < k$ so $k$ is not a lower bound.
So $\min G$ is the $\inf G$.
