Limit Point of a Subset using Open Ball

137 Views Asked by At

I am studying analysis II and I have come across the following lemma:

Lemma: Let $(V, ||.||)$ be a normed space, $E$ any subset of V. Then a point $\vec{y} \in V$ is a limit point of $E$ iff $(B_{r}(\vec{y})\backslash \{\vec{y}\})\cap E \ne \emptyset$ for every $r$, where $B_{r}(\vec{y}) = \{ \vec{x} \in V: ||\vec{x} - \vec{y}|| < r \}$.

I have understood the proof for this lemma but intuitively I still have some problems with it. For example let $V = \mathbb{R}$ and $E = (-1,1)$. The limiting points of this set are $-1$ and $1$. Now, if we let $\vec{y} = 0$ and $r > 0$, $(B_{r}(\vec{0})\backslash \{\vec{0}\})\cap E \ne \emptyset$ as the intersection will include some part of $E$ which would mean $0$ is a limiting point as well (this reasoning can be applied to every point inside the set $E$). Clearly, there is something wrong with my reasoning but I can't seem to find it. Help would be much appreciated.