Question
We have a set E which is a subset of the real numbers. There is a sequence ${x_n}$ such that $\{x_n\} \subseteq E$. Suppose there is another sequence $\{y_n\}$ such that the limit as $n$ goes to infinity for both sequences is the number $y$.
We also assume that every term in the sequence $\{y_n\}$ is an upper bound for $E$. Show that $y$ is an upper bound for $E$.
My reasoning
First since the sequence $\{x_n\}$ is inside of E, that means that every term of the sequence must be an element of the set E. Since all terms in the sequence $\{y_n\}$ are upper bounds for the set $E$, then this means that $\{x_n\} \leq \{y_n\}$ for all $n$.
I'm really stuck though. I have a feeling that that both sequences are converging to the sup of the set E, but i'm not really sure how to get started with this. A few hints or things to consider would be helpful.
Best Answer
Maybe it is better if you try this by contradiction.
Suppose that $y$ is not an upper bound for $E$. What does this mean? It means we can find some $s \in E$ such that $y < s$. But if you draw the real number line, and mark $y$ somewhere to the left of $s$, think about any small open interval around $y$ that doesn't contain $s$.
For example, if we let $\epsilon = |y - s|$ (i.e., $\epsilon$ is the distance between $y$ and $s$), then think about $(y - \frac{\epsilon}{3}, y + \frac{\epsilon}{3})$.
What does it mean for the sequence $y_{n}$ to converge to $y$? It means that for any open interval around $y$, there is some $N$ such that for all $n \geq N$, $y_{n}$ is in that interval. But that means there must be some $N$ such that for all $n \geq N$, $y_{n}$ is in $(y - \frac{\epsilon}{3}, y + \frac{\epsilon}{3})$. But that means we can find terms in the $\{y_{n}\}$ sequence that are in this interval. But every number in this interval is less than $s$, so we can find terms in the sequence that are less than $s$. This contradicts that all of the terms in the $\{y_{n}\}$ sequence are an upper bound for $E$.