How do I prove I found every element in the closure of $A$? Suppose we have this interval $A=]1, 5].$ We know that $A$ is in the closure of $A$.
If we add an arbitrary number $b > 0$ to $1$ we get: $1 + b > 1$, so the intersection of the sphere around $1$ with radius $b$ with $A$ is non-empty. So $1$ is in the closure of $A$, and the closure of $A$ is $[1,5]$.
Do I need to prove that that are no other terms in the closure of $A$? If so, how can I do this?
 A: Since you appear to be at the level of elementary analysis, it seems a good idea to prove it. It is not very difficult. You already understand about the key ideas of "arbitrarily small positive number" $b$, and then using $b$ as a radius of a little ball.
If you pick an arbitrary real $t$ such that $t<1$ or $t>5$, you can prove that $t$ is not in the closure of $A$. This you can do by specifying a radius $b=b(t)$ so that the sphere around $t$ of radius $b$ satisfies $\_\_\_\_(??)\_\_\_\_.$
Can you say what the missing condition (??) must be? If not, try the specific value $t=7$, and draw a little picture.
A: Take any point $x \not\in [1,5]$, and show that $x$ is not a limit point of $A$. Suppose $x < 1$ (the case $x > 5$ is symmetric).
Now let $\epsilon = \dfrac{1-x}{2}$ and you have found an open neighbourhood $(x-\epsilon, x+\epsilon)$ which does not intersects with $A$, which by definition implies that $x$ is not a limit point of $A$.
Therefore, you found all limit points of $A$ and $[1,5]$ is its closure.
