I am reading a paper and there is a Lemma stating that:
Let $f : X \rightarrow X$ be continuous, where $X$ is a non-empty Hausdroff topological space and suppose that every backward invariant subset of $X$ with nonempty interior is dense. Then $f(X)$ is dense.
In the proof, the author claimed that $G=X \setminus \overline{f(X)}$ cannot be the set of one point.
The author said as followings.
We can not have $G=\left \{ x \right \}$ since this would give $X= \bar{G}=\overline{\left \{ x \right \}} = {x}$ using the fact that $X$ is Hausdroff, which is impossible since $\overline{f(X)} \neq \varnothing$ and $\left \{x \right \} \cap \overline{\left \{ f (X) \right \}} = \varnothing$.
I do not understand how we deduce that $G \neq \left \{x \right \}$ since the property of Hausdroff of $X$.
Could you please show me in more detail?
Thank you for your time.
First notice that $G$ is open in $X$, as it is the complement of a closed set. It's also backwards invariant, since its preimage is empty by construction. Then given that $G$ is nonempty, the hypothesis of the lemma implies that $G$ must be dense.
Now assume that $G$ is a singleton $\{x\}$. We have proved that $X = \overline{G} = \overline{\{x\}}$ as stated in the first line, and this equals $\{x\}$ because singletons are closed in Hausdorff spaces. So $X$ would have to be a singleton set itself, and you can check (e.g. as in the second line) that even this doesn't work.