If $\mathcal{B}_{0}=\{Int(B):B\in\mathcal{B}\}$ is a local base of a point $x,$ then $\mathcal{B}$ is also a local base of $x?$

Question

I know that, if a collection $\mathcal{B}$ is a base for a point of $X,$ then $\{Int(B):B\in\mathcal{B}\}$ is local base for such point too, but what can we say about the converse? I think this could be false. so I'm trying to give a counterexample of the next:

If $\mathcal{B}$ is a collection of subsets of a topological space $X$ such that $\mathcal{B}_{0}=\{Int(B):B\in\mathcal{B}\}$ is local base of the point $x\in X,$ then $\mathcal{B}$ is also a local base of $x.$

The definition of local base of a point that I know is the next:

Given a topological space $(X,\tau),$ a family $\mathcal{B}(x)$ is local base of $x$ if every element of $\mathcal{B}(x)$ is a neighborhood of $x$ and, for each neighborhood $\mathcal{U}_{x}$ of $x,$ there is $V\in\mathcal{B}(x)$ such that $V\subset\mathcal{U}_{x}.$

So, my feeling is the proposition is false, because $\mathcal{B}$ is a family of neighborhoods for $x,$ but the second condition of the definition can't be satisfied because we don't know the structure of $\mathcal{B},$ but I cannot give a concrete counterexample.

Any idea is welcome.

Answer 1

Let $X = \{0,1,2\}$ with the "boring" topology $\{\emptyset, \{0\}, X\}$. (As an aside: this boring three point space has the property that any topological space $P$ can be embedded topologically into some power $X^I$, so it's universal in some sense.)

Then $\mathcal{B} = \{\{0,1\}\}$ has the property that it is not a local base at $0$ but that $\{\operatorname{int}(\{0,1\})\} = \{\{0\}\}$ is a local base at $0$. So this is a possible counterexample.

Answer 2

For all x, let $B_x$ be a local base for x.

Try this.
If for all a,b,x, U in $B_a$, V in $B_b$, x in U $\cap$ V,
some W in $B_x$ with W subset U $\cap$ V,
then $\cup_x B_x$ is a base.

That is a clue how to make a counter example.