Proving that a set has a minimum

Question

Let's suppose $(a_n)$ is a sequence of real numbers, where $n$ belongs to natural number and $a_n$ diverges to $\infty$ as $n\to \infty$. Let $A$ be a set which consists of all the terms of the sequence.

How do I prove that the set $A$ has a minimum?

This seems unusually hard to me. Since sequence must have the smallest number which belongs to the set, the set must have a minimum. But how do I mathematically prove this? Where do I start? Just a hint would be so useful.

Answer 1

For starters, let's set up the definition of the fact that $a_n\to+\infty$ as $n\to+\infty$:

For any $M\in\mathbb{R}$ there exists an index $N\in\mathbb{N}$ such that for all $n\ge N$: $a_n\ge M$.

Now, we can start with the observation that $a_1\in A$ by definition of the set $A$. The definition above applied to $M=a_1$ tells us that there exists an index $N\in\mathbb{N}$ such that for all $n\ge N$: $a_n\ge a_1$. Now the value of $\color{blue}{m=\min\{a_1,a_2,\ldots,a_{N-1}\}}$ will be the answer, because any other element of the sequence, i.e. any element $a_n$ where $n\ge N$, would have a larger value: $a_n\ge a_1\ge m$.

Answer 2

If $a_n$ goes to infinity, then exists $m\in\Bbb N$ such that $a_m\le a_N$ for all $N>m$. Now you have construct new set, $B$ which includes $a_0,a_1,\cdots,a_m$, because $B$ is finite it had minimum, and because every element that is in $A$ and not on $B$ is less than an element in $B$ the minimum of $B$ is the minimum of $A$

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This is assuming that when you say diverges to infinity you mean positive infinity (I saw cases where it wasn't the case)

Answer 3

You could perhaps use some point-set topology here:

Roughly speaking, look at your set, and do a cut-off so that it is a closed and bounded set - ignore the tail-end of the sequence that diverges to infinity. (You have to work out the specifics here ... )

With this set, you have sequential compactness - every convergent sequence in this set converges to a point that is in the set. By definition, your set is closed. As the set is also bounded, by the extreme value theorem, your closed and bounded set achieves a minimum and maximum.