Finding a topology doesn't convey Hausdorff property by an injective continuous function

Question

Is there a topology on $\mathbb{R}$ that for which , set $A=(0,1)$ be Hausdorff and $B=[0,1]$ not to be Hausdorff with subspace topology and there exist an injective continuous function $f:A \to B$ ?

Answer 1

Yes. Consider the topology $\tau$ whose elements are the open subsets of $(0,1)$ (with respect to the usual topology) and the unions of an open subset of $(0,1)$ (again, with respect to the usual topology) with $\mathbb{R}\setminus(0,1)$. Then $(0,1)$, as a subspace of $(\mathbb{R},\tau)$, has its usual topology and therefore it is Hausdorff. But $[0,1]$ isn't, since every open set which contains $0$ also contains $1$ (and vice-versa).

Now, let $f\colon A\longrightarrow B$ be the function defined by $f(x)=x$. It is injective and continuous.