Show that $a$ is limit point to the set $[a,b]$ I'm trying to prove that, in the real line $\mathbb R$ with the standard topology, the point $a$ is a limit point of the interval $[a,b]$.
For every neighborhood $U$ of $a$, we have to show that exist some real $y\neq a$ in the intersection $U \cap [a,b]$. Let $B_{\rho}$ be the ball contained in $U$, of radius $\rho \in \mathbb{R}_{>0}$; our $y$ should satisfy the two following conditions: $a - \rho \leq y \leq a + \rho$ and $a \leq y \leq b$. We know that in a ordered field we can always pick a number between two elements of the set, so let
$$y = \frac{\frac{a - \rho}{2} + \frac{a + b + \rho}{2}}{2}$$
Now, $y$ should be $\in U \cap [a, b]$, but it's evident that this is not true for every value of $\rho$, $a$ and $b$.
I'm not sure if this is the proper way to do this. I thought that maybe there is a way that uses $\operatorname{min}$, but I'm not sure about it.
Note: I'm not using the definition of limit point that involves limit of sequences.
 A: You are thinking way too hard. 
Any ball around $a$ is the interval $(a-\epsilon, a +\epsilon)$ for some $\epsilon > 0$.  And for any $\epsilon; 0 < \epsilon < b-a$ then $(a-\epsilon, a +\epsilon)\cap [a,b] = [a,a+\epsilon)\subset [a,b]$ which consists of infinitely many $y; a< y < a + \epsilon\le b$ so that $y \in [a,b]$.
.......
If you feel you can't simply claim there are infinitely many points in $(a,a+\epsilon)$ (and you totally can) but feel a need to specifically identify one of them; then $\frac {a + (a + \epsilon)}2 = - a + \frac \epsilon 2$ will do, but hopefully, you see that is just mechanical busy work.
A: Let $U$ be a neighbourhood of $a$. Therefore, there exists an open set $x \in V \subseteq U$, and in particular, an open ball around $a$ contained in it, $(a-\varepsilon,a+\varepsilon) \subseteq U$. By archimeadianity there exists $n \in \mathbb{N}$ such that $n > \frac{1}{\varepsilon}$ and $n > \frac{1}{b-a}$, so that $a+\frac{1}{n} < b$ and $a+\frac{1}{n} < a+\varepsilon$. Thus,
$$
a+ \frac{1}{n} \in (a-\varepsilon,a+\varepsilon) \cap [a,b]
$$
This avoids the problem of taking the midpoint and checking if it lies on the desired set.
