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Essentially I need feedback, mostly on writing style and accuracy and tightness of arguments.

Suppose $\mathcal{B}$ converges to $x$. For every open neighbourhood $U(x)$ of $x$ in $X$ there exists $B_j \in\mathcal{B}$ such that $B_j\subset U(x)$. Clearly, $B_j \cap U(x) \neq \emptyset$. We have to show that $B_i \cap U(x) \neq \emptyset$ for all $i \in I$ such that $i \neq j$. Since for all $i,j \in I$ there exists $k \in I$ such that $b \in B_k \subset B_i \cap B_j$, therefore $B_i \cap B_j \neq \emptyset$ for all $i,j \in I$. It follows that $B_i \cap U(x) \neq \emptyset$ for all $i \neq j \in I$. Thus, for every $U(x) \in \mathcal{T}$ and every $B_i \in \mathcal{B}$, $U(x) \cap B_i \neq \emptyset$.\ Suppose $(X, \mathcal{T})$ is a Hausdorff space. Consider a filterbase $\mathcal{B}$ in $X$ converging to $x\in X$.We show that, if $y \in X$ and $y\neq x$, then $\mathcal{B}$ cannot accumulate at $y$. Since, $x\neq y$ and $X$ is Hausdorff, there exist open neighbourhoods $U(x) \in \mathcal{T}$ and $V(y) \in \mathcal{T}$ of $x$ and $y$ respectively such that $U(x) \cap V(y) =\emptyset$. Since $\mathcal{B}$ converges to $x$, there exists a filterbasis element $B_i\in \mathcal{B}$ for every open neighbourhood $U(x)$ of $x$ such that $B_i \subset U(x)$. It follows that there exists $B_i \in \mathcal{B}$ such that $B_i \cap V(y)= \emptyset$. Therefore, $\mathcal{B}$ cannot accumulate at $y$.

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$I$ is defined in the definition of filterbase as follows: A filterbase $\mathcal{B}$ in $X$ is a family of sets $\mathcal{B}=\{B_i \mid i \in I \}$. –  Fatsho Feb 20 '13 at 12:12

1 Answer 1

I would do it somewhat differently:

Suppose we have a filterbase $\mathcal{B} = \left\{B_i : i \in I \right\}$, and $x \in X$. Then $\mathcal{B}$ accumulates at $x$ iff

$$\forall i \in I, \forall \mbox{ neighbourhoods } U(x) \mbox{ of } x, U(x) \cap B_i \neq \emptyset$$

or equivalently $x$ is in the closure of all $B_i$. Now, if $\mathcal{B} \rightarrow x$, then $x$ is an accumulation point: pick an arbitrary $U(x)$ neighbourhood of $x$, and an arbritary $B_i$ in $\mathcal{B}$. Then by convergence for some $j$ we have $B_j \subset U(x)$ and also we have by the definition of a filterbase, applied to $B_i$ and $B_j$ some $B_k$ with $\emptyset \neq B_k \subset B_i \cap B_j \subset B_j \subset U(x)$. It follows that indeed $U(x)$ and $B_i$ intersect, as required.

Now suppose $X$ is Hausdorff, $\mathcal{B} \rightarrow x$ and $y$ is an accumulation point of $y$. Suppose $y \neq x$. Then there are disjoint neighbourhoods $U(y)$ and $U(x)$ of $y$ resp. $x$. By convergence, for some $i$, $B_i \subset U(x)$. But if $y$ is an accumulation point $U(y)$ should intersect $B_i$, which cannot be. Contradiction. So $y = x$ and $x$ is the unique accumulation point of $\mathcal{B}$.

So essentially you got the arguments right, I think.

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