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Let $A\subset\mathbb R$. Show for each of the following statements that it is either true or false.

  1. If $\min A$ and $\max A$ exist then $A$ is finite.
  2. If $\max A$ exists then $A$ is infinite.
  3. If $A$ is finite then $\min A$ and $\max A$ do exist.
  4. If $A$ is infinite then $\min A$ does not exist.

My attempts so far:

  1. This statement is wrong. Let $A=[a;b]\cap\mathbb Q\subset\mathbb R$ with $a<b$. It is obvious that $\min A=a$ and $\max A=b$. Assume now, that $A$ is finite, then we would be able to enumerate every element in $A$, however $\mathbb Q$ is dense in $\mathbb R$ and therefore we can find for each $x_k,y_k\in A$ a $r_k\in A$ in $[x_k;y_k]$, such that $x_k<r_k<y_k$. E.g. this can be achieved with the arithmetic mean $r_k=(x_k+y_k)/2$. Starting with $[x_1=a;y_1=b]$ one could create infinitely nested intervals $[x_k;r_k]$ and can therefore find infinite elements in constrast to the assumption, that $A$ is finite.

  2. This statement is wrong. Let $A=(0;n]\cap\mathbb N\subset\mathbb R$ with $n\in\mathbb N$ and therefore $A$ is bounded by $n$ with $\max A=n$. Assume $A$ would be infinite; with $|A|=n-1$ follows, that $A$ must be finite contrary to the assumption that it is infinite.

  3. ???

  4. This statement is wrong. Let $A=\mathbb N\subset\mathbb R$ and we know that $A$ is infinite. However $A$ has a lower bound and even a minimal element where $\min A=1$ in constrast to the assumption that it does not exist.

I would like to know whether my attempts for (1), (2) and (4) are plausible or incomplete. Furthermore I need some hints on how to prove (3).

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up vote 1 down vote accepted

What you’ve written for (1) isn’t quite right. Suppose that $a=\sqrt2$ and $b=\sqrt2+1$; then $[a,b]\cap\Bbb Q$ has neither a maximum nor a minimum element, since $a$ and $b$ are irrational. In any case you’re working much too hard. Why not just take $A=[0,1]$? Then $\min A=0$, $\max A=1$, and since $\frac1n\in A$ for every $n\in\Bbb Z^+$, $A$ is clearly infinite.

(2) is okay, though again you’re working harder than necessary: just take $A=\{0\}$, say. Then $\max A=0$ certainly exists, and $A$ is certainly finite!

(4) is fine.

(3) is almost true: it’s true if $A$ is non-empty. One way to prove it is to show that if $A$ does not have a maximum (resp. minimum) element, then $A$ is not finite. Suppose that $A$ has no maximum element. Then there is some $a_1\in A$ such that $a_1\ge 1$. Given $a_1,\dots,a_n\in A$, there is an $a_{n+1}\in A$ such that $a_{n+1}\ge\max\{\dots\text{what?}\dots\}$, and in this manner we recursively build an infinite subset of $A$.

I deliberately avoided the argument that if $A\ne\varnothing$ is finite, we can simply index it as $\{a_1,\dots,a_n\}$ for some $n\in\Bbb Z^+$ in such a way that $a_1<\ldots<a_n$; depending on just how rigorous you want to be, this might feel like hand-waving. You could, however, legimitately say that $A=\{a_1,\dots,a_n\}$ for some enumeration and then argue as follows. Let $b_1=a_1$. If $n>1$, let $b_2=\min\{b_1,a_2\}$. In general, given $b_k$ for some $k<n$, let $b_{k+1}=\min\{b_k,a_{k+1}\}$. An easy induction shows that for each $k\le n$ we have $b_k\le a_i$ for $i=1\dots,k$, so in particular $b_n\le a_i$ for $i=1,\dots,n$, and therefore $b_n=\min A$. A similar argument shows that $\max A$ exists.

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(1) As mentioned in fgp's answer I will reuse your idea. Quite easy, when being reminded of this fact with $1/n$. (2) I like $A=\{1\}$ more, but yeah quite complex what I did there. (3) I'm sorry, but I didn't really get everything you wanted to explain. fgp's description was more intuitive on the first look. (4) Ok :) – Christian Ivicevic Oct 20 '12 at 16:16

For (1) your argument seems more complex than necessary. $A = [a,b]$ alone is a sufficient counterexample. It obviously has a minimum and a maximum, yet its not finite. Should you think that it's not obvious that $A$ is infinite, observe that $f:\, \mathbb{N} \to A,\, n \to a + \frac{b-a}{n}$ is an injective map from $\mathbb{N}$ to $A$.

For (2), $A = \{1\}$ is a sufficient counterexample. Again, there can quite obviously be no injective map from $\mathbb{N}$ to $A$.

For (3), simply argue that it's trivial to compute the actual minimum and maximum. You only have to test finitely many elements, after all. More formally, let $f(1) = a_1 \in A$, let $f(2) = a_2 \in A$ with $a_2 > a_1$ and so on. If you cannot continue at some point, you've found a maximum. If you can continue arbitrarily long, you've found an injective $f:\, \mathbb{N} \to A$, i.e. $A$ is infinite. Similarly for the minimum.

For (4), I like your proof. Another possibility is taking $A = [a,b]$ again. We already know that $A$ has a minimum and is infinite from (1). In fact, (4) follows from (1). In (1) you showed that $$ \forall A (\lnot(\text{MinExists}(A) \land \text{MaxExists}(A) \implies \text{Finite}(A))) $$ which is equivalent to $$ \exists A (\lnot\text{Finite}(A) \land \text{MinExists}(A) \land \text{MaxExists(A)}) $$ which contradicts $$ \lnot\exists A (\lnot\text{Finite}(A) \land \text{MinExists}(A)) $$ and thus $$ \forall A(\text{Finite}(A) \lor \lnot\text{MinExists}(A) $$

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(1) I do not want to use injective maps as proof though I like your idea. However I will use Brians general approach with $\forall x\in\mathbb R,x>0:\exists n\in\mathbb N:0<1/n<x$. (2) My approach is waaaaay to complex :D (3) Good idea going through all elements and comparing them with a finite number of operations. I will reuse this one. (4) Thanks, however my idea was really easy and nothing special. – Christian Ivicevic Oct 20 '12 at 16:13

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