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

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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).