Let $A\subset\mathbb R$. Show for each of the following statements that it is either true or false.
- If $\min A$ and $\max A$ exist then $A$ is finite.
- If $\max A$ exists then $A$ is infinite.
- If $A$ is finite then $\min A$ and $\max A$ do exist.
- If $A$ is infinite then $\min A$ does not exist.
My attempts so far:
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.
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.
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).