I am not sure if I understand you. The point is not an element of the open interval . Furthermore, if , then so and similarly if , then so . This means that really is the only point of in the open interval .
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 as your question seems to suggest. In fact, for any two (different) real numbers, , one can always find an open neighbourhood of which does not contain . Trying to prove this is a good exercise but if you struggle I (or any textbook) can tell you why this is the case.
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 as your question seems to suggest. In fact, for any two (different) real numbers, , one can always find an open neighbourhood of which does not contain . Trying to prove this is a good exercise but if you struggle I (or any textbook) can tell you why this is the case.