Proving bounded sequence in $\mathbb R$ which has just one limit point is convergent My approach to proving this as follows :
What would be your advice on me? Is my proof incorrect/correct, what else should be mentioned etc? I'm just new to analysis in mathematics.
So let $a_n$ be this sequence. We are in $\mathbb R$ so by the Least upper bound property $\inf(a_n)$ and $\sup(a_n)$ exist. Either $\sup(a_n)$ or $\inf(a_n)$ will be achieved as $n$ goes to infinity. Let's say $a_n\to L$ as $n$ goes to infinity. We know this should either be $\sup(a_n)$ or $\inf(a_n)$ as I mentioned above. If let's assume $a_n$ increases to this would be $\sup(a_n)$
Therefore $$\sup(a_n)=L$$
Let's say that there exists another limit $M$ For $M$ to be different than $L$ $M=\inf(a_n)$ should be the case.
But since we assumed $a_n$ is an increasing function, $|a_n-\inf(a_n)|<|a_{n+1}-\inf(a_n)|$ so as $n$ goes to infinity it is impossible for us the reach $\inf(a_n)$ therefore convergence happens only when limit point exists.
If more than two limit points exist, this would show us that $a_n$ is not monotonic, therefore can't converge.

There are several issues in what you said:


*

*What do you mean by "Either $\sup_n a_n$  or $\inf_n a_n$ will be achieved as $n$ goes to infinity"?

*$\sup_n a_n$ and $\inf_n a_n$ are not related to the limit points of the sequence $(a_n)$. Example: $a_n = -1, 1, 0, 0,0,\dots $. $\inf_n a_n=-1$, $\sup_n a_n=1$ are not limit points of the sequence $(a_n)$.

*There can be no subsequence of $(a_n)$ converging to $L = \sup_n a_n$, cf. previous example.


More interesting would be to consider $\liminf_n a_n$ and $\limsup_n a_n$, and to prove that those numbers are equal.
