Willard's very extreme example of topological space

Question

In Willard's General Topology, page 32, he says:

"If $y$ is close to $x$ in a metric space, then $x$ is close to $y$; but it can happen in a topological space that $y$ is in every neighborhood of $x$ while $x$ is in no neighborhood of $y$ (a very extreme example; this doesn't happen in useful topological spaces, although many useful spaces lack symmetry in some degree)"

But I do not get how could this be possible. How can such extreme example of topological space exist if the whole space contains itself and then it is a neighborhood of any point, so there cannot be a point that is in no neighborhood of the other?

Any clarification on what he meant would be accepted, or even an explicit example of such very extreme topological space.

Thanks.

Answer 1

You are correct: the entire space will always be a neighborhood of $y$ which contains $x$. (I'm assuming $x \neq y$.) I'm guessing he wanted to exclude that neighborhood as a trivial example, and to say that no other neighborhoods of $y$ contain $x$.

In the Sierpiński space, for example, $1$ has a neighborhood that does not contain $0$, but $0$ has a single neighborhood, which does contain $1$. More generally, the "excluded point topology" will give you more examples.