Let $(X,\Gamma)$ a topological Hausdorff space and $A\subset X$. Show that $x$ is an "accumulation point" of $A$ if and only if every neighborhood of $x$ contains infinitely points of $A$.
I proved that a sequence in Hausdorff space cannot converge to more than one point, any hint thanks!
Best Answer
Suppose that $x$ is an accumulation point of $A$ and $U$ a neighborhood of $x$. Since $x$ is an accumulation point of $A$, by definition, there exists a point $a_0\in A\cap U$ such that $a_0$ is distinct of $x$.
Suppose constructed $a_n$ such $a_n\in U$, $a_0,...,a_n,x$ are pairwise distinct. Since $X$ is separated, there exists a neighborhood $U_n$ of $x$ such that $a_i$ is not in $U_n$ for $i=0,...,n$, $U_n\cap U$ is a neighborhood of $x$, since $A$ is a point of accumulation of $A$, there exists $a_{n+1}\in U_n\cap U\cap A$ distinct of $x$, the sequence of points $(a_n)$ is contained in $U$ and $a_i\neq a_j$ if $i\neq j$.
Suppose that every neighborhood of $x$ contains an infinite number of elements of $A$. Let $U$ be a neighborhood of $A$, since $A\cap U$ is infinite, there exists $a\in A\cap U$ distinct of $x$, henceforth $x$ is an accumulation point of $A$.