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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 $\inf G = x \in G$.

user10354138
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npdfe
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3 Answers3

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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 $G$.

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    "Since G is nonempty and finite, its minimum minG exists. " Why? "all you need to show is that minG=infG." If $\min S$ exist then $\min$ = \inf S$. I think the heart of this question is proving every finite set of real has in minimum element. – fleablood Aug 09 '20 at 00:30
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    This answer seems correct to me. – Rachid Atmai Aug 09 '20 at 00:34
  • @fleablood I pretty much had the same 'why?' question as you did. It's easy to see how people can disagree with what exactly a proof consists in, when not everyone is convinced. – Doug Spoonwood Aug 09 '20 at 00:42
  • Sorry, so how do you conclude this proof? – npdfe Aug 09 '20 at 00:45
  • @npdfe: check the update. –  Aug 09 '20 at 00:47
  • "This solution assumes that you are allowed to use the following fact" I have no idea why you would assume the user would be allowed to assume that. It's not an axiom. – fleablood Aug 09 '20 at 00:50
  • @fleablood: You can prove it if you like. We must assume something in order to give an answer. Otherwise, we would be forced to begin with the ZFC axioms and write a book for an answer. –  Aug 09 '20 at 00:52
  • @user798202 thanks you! – npdfe Aug 09 '20 at 01:32
  • @npdfe: you are welcome! –  Aug 09 '20 at 01:33
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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.

Doug Spoonwood
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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 $\inf G$.

fleablood
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