Neighborhood Base (the definition) In Steven G. Krantz' A Guide To Topology, a countable neighborhood base is defined:
Let $(X,U)$ be a topological space.  We say that a point $x\in X$ has a countable neighborhood base at $x$ if there is a countable collection $\{U_{j}^{x}\}_{j=1}^{\infty}$ of open subsets of $X$ such that every neighborhood $W$ of $x$ contains some $U_{j}^{x}$.
Here is a link.
Now, to define a neighborhood base at $x$, the obvious thing to do would be to simply drop the countable requirement, and replace $\{U_{j}^{x}\}_{j=1}^{\infty}$ with some collection $\{U_{\alpha}^{x}\}_{\alpha\in J}$ with index set $J$.
My question:  (and I've seen this same definition in other books)  Why do we not require that each $U_{\alpha}^{x}$ contain the point $x$?  My intuitive idea of what a neighborhood base ought to be completely falls apart without this requirement.  All examples of neighborhood bases seem to satisfy this.  Is it a consequence of the definition?
Is there a better way to think about neighborhood bases?
 A: As was mentioned by a couple people in the comments, the answer is that this is neither the usual nor a particularly good definition of a neighborhood base.
In fact, let $X$ be any topological space, let $x\in X$, and define $\mathcal{B}_x:=\{ \emptyset \}$.  Then certainly, for any neighborhood $W$ of $x$, there is some element of $\mathcal{B}_x$ which is a subset of $W$.  This, however, is silly.
Suppose, however, that we even amended the definition you gave to require that elements of a neighborhood base were all nonempty.  Even in this case, the definition is lacking.
One way to see this is because, for the usual definition, we have that, if each $\mathcal{B}_x$ is a neighborhood base at $x$ for all $x\in X$, then $U\subseteq X$ is open iff for every $x\in U$ there is some $B_x\in \mathcal{B}_x$ such that $B_x\subseteq U$.  With even the amended definition of what you gave (which hereafter I shall refer to simply as "your definition"), this is no longer so.
For example, take $X:=\{ 0,1,2\}$ with the topology $\{ \emptyset ,\{ 0\} ,X\}$, and define $\mathcal{B}_0,\mathcal{B}_1,\mathcal{B}_2$ all to be $\{ \{ 0\} \}$.  Then, each $\mathcal{B}_x$ is a neighborhood base according to your definition.  But now it is the case that for every element $x\in U:=\{ 0,1\}$ there is some element $B_x\in \mathcal{B}_x$ such that $B_x\subseteq U$, even though $U$ is not open!
