topological space satisfying the first axiom of countability If a topological space $(X,\tau)$ satisfies the first axiom of countability, then for every $x\in X$, there exists a countable local base $\{B_k\}$ at $x$. However, I find in many proofs the local base will be further assumed to satisfy $B_{k+1}\subset B_k$ for every $k\in N$. Why is this assumption correct?
Thanks!
 A: By definition, a countable neighborhood basis of a point $x \in X$ is a filter base for the neighborhood filter $\mathcal{V}(x)$ which is countable. Let $\{B_k\}$ be this filter base. Then $\{\cap_{i=1}^k B_i\}$ is also a filter base for the neighborhood filter and is also countable:
Since all the $B_k$ are neighborhoods of $x \in X$, any intersection is non-empty i.e. $\emptyset \not\in \{\cap_{i=1}^k B_i\}$. 
Let $N = \cap_{i=1}^n B_i$ and $M = \cap_{i=1}^m B_i$. Then $\cap_{i=1}^{\max{m,n}} B_i \subseteq N \cap M$.
Hence it is a filter base.
Let $V \in \mathcal{V}(x)$ be arbitrary, then there exists a $k \in \mathbb{N}$ s.t. $B_k \subseteq V$, and thus $\cap_{i=1}^k B_k \subseteq V$.
Hence it is a filter base for the neighborhood filter.
Hence if for every point $x \in X$ there exists countable neighborhood base $\{B_k\}$, then there exists a countable neighborhood base with $B_{k+1} \subseteq B_k$.
A: You can replace each open set with its intersection with the (finitely many) preceding ones to get such a local basis. 
A: If $(B_n)_{n \in \omega}$ is a local base at $x$, then define for all $n \in \omega$:
$$B'_n = \bigcap_{k \le n} B_k$$
and then all $B;_n$ are still open neighbourhoods of $x$ (as finite intersections of open neighbourhoods of $x$) and $B'_{n+1} \subseteq B'_n$ is obvious from the definition, and $(B'_n)_{n \in \omega}$ is still a local base at $x$:
If $O$ is open and contains $x$, for some $m\in \omega$ we will have $(x \in )B_m \subseteq O$, as we started with a local base at $x$, and by definition $B'_m \subseteq B_m$ so also $(x \in )B'_m \subseteq O$ and we're done.
The countability ensures that all the intersections we take are finite, and we cannot always get a decreasing neighbourhood base for larger cardinalities of local bases. But if there is a countable one, there is a countable decreaasing one. The advantage of having a decreasing one is that is we pick $x_n \in B'_n$ for each $n$ we are guaranteed that $x_n \to x$, because the tail requirement in sequence convergence will be automatic..
