$\mathbb R$ with usual topology is a Hausdorff space. By the definition of Hausdorff space, for any two distinct point in $\mathbb R$ we can find disjoint neighbourhoods… Now consider the set $A$=$\{{{1}\over{n}}\}$ where $n\in \mathbb N$. We know that the sequence $\{{{1}\over{n}}\}$ converges to "$0$". Clearly $0$ does not belong to $A$, that is every point in $A$ and the point $0$ are distinct. Hence by definition of Hausdorff space we can find a neighbourhood of $0$ which does not intersect $A$. This shows that $0$ is not a limit point. This is contradiction to ${{1}\over {n}}\rightarrow 0$. Is my opinion is correct? please spot where i make mistake?
my thought: consider every point in $A$ and the point $0$, we can get disjoint neighbourhoods because they are disjoint. In this way of thinking I am getting lot of complications.. what can I do?
[Math] Is $\mathbb R$ with usual topology a Hausdorff space
general-topology
Best Answer
You didn't take a point, you took a sequence of points. That's not just one point, that's infinitely many points.
Each $\frac1n$ and $0$ can be easily separated by the intervals $(-1,\frac1{4n})$ and $(\frac1{2n},2)$.