Mathematics III - Foundations of Analysis - Summer 24

I am not sure if I understand you. The point $a+ \frac{1}{n+1}$ is not an element of the open interval $U = (a+ \frac{1}{n+1}, a+ \frac{1}{n-1} )$. Furthermore, if $k \geq n+1$, then $a+ \frac{1}{k} \leq a+ \frac{1}{n+1}$ so $a+ \frac{1}{k} \notin U$ and similarly if $k \leq n-1$, then $a+ \frac{1}{k} \leq a+ \frac{1}{n+1}$ so $a+ \frac{1}{k} \notin U$. This means that $a+ \frac{1}{n}$ really is the only point of $E$ in the open interval $U$.

If this does not answer your question could you please specify what you mean by "the Neighborhood (a + 1/n)"? Every point has an infinite number of neighbourhoods and it is not true that all of them contain the point $a+ \frac{1}{n+1}$ as your question seems to suggest. In fact, for any two (different) real numbers, $x, y$, one can always find an open neighbourhood of $x$ which does not contain $y$. Trying to prove this is a good exercise but if you struggle I (or any textbook) can tell you why this is the case.