Let $(X, \tau)$ be a topological space. We say that $(X,\tau)$ is $T\_{\text{inj}}$ if there is an injective map $f:X\to\tau$ such that $x\in f(x)$ for all $x\in X$.


It is not hard to see that $T\_1$ and all the stronger separation axioms imply $T\_{\text{inj}}$.


**Questions.** Does $T\_0$ imply $T\_{\text{inj}}$? Is there any implication between $T\_{\text{inj}}$ and other weak separation axioms?



Existence of injective neighborhood selection function as separation axiom
$\textbf{Finite $T\_{0}$-spaces are $T\_{inj}$}$


I claim that every finite $T\_{0}$-space is $T\_{inj}$. If $(X,\tau)$ is a finite $T\_{0}$-topology, then there exists a partial ordering $\leq$ on $X$ so that the open sets are precisely the upwards closed sets. Therefore, let $f:X\rightarrow\tau$ map the element $x$ to the smallest upwards closed set containing $x$. Then $f$ is an injective mapping. 


$\textbf{Not every $T\_{0}$-space is $T\_{inj}$}$


Note: Due to a remark by Eric Wofsey, I am generalizing the counterexample I had earlier. Suppose that $X=A\cup(\omega+1)$ and $A$ is disjoint from $\omega+1$. Give $X$ any $T\_{0}$-topology $\mathcal{T}$ such that if $\alpha\in\omega+1$, then the neighborhoods of $\alpha$ are the sets of the form $\{x\in\omega+1|x>\beta\}\cup A$ for some $\beta<\alpha$ and the set $X$ itself. I claim that $X$ is not $T\_{inj}$. Suppose that $f:X\rightarrow\mathcal{T}$ is injective and $x\in f(x)$ for each $x\in X$. Then since the only neighborhood of $0$ is $X$ we conclude that $f(0)=X$. I now claim that if $n$ is a positive integer, then $f(n)=\{x\in\omega+1|x\geq n\}\cup A$. We shall proceed by induction. If $n=1$, then the only neighborhoods of $f(1)$ are $X$ and $X\setminus\{0\}$. However, since $f$ is injective, we have $f(1)=X\setminus\{0\}$. Now assume that $n>1$. Then $f(n)\in X,X\setminus\{0\},...,X\setminus\{0,...,n-1\}$. However, since $f(i)=X\setminus\{0,...,i-1\}$ for $0<i<n$ and $f(0)=X$ and $f$ is injective, we have $f(n)=X\setminus\{0,...,n-1\}$. However, the only neighborhoods of $f(\omega)$ are the neighborhoods $X$ and $X\setminus\{0,...,n\}$ and $f(n)=X\setminus\{0,...,n-1\}$ and $f(0)=X$, so $f(\omega)=f(n)$ for some $n<\omega$. Therefore the function $f$ is not injective.


In particular, if $\lambda$ is an ordinal greater than $\omega$ and $\lambda$ is given the topology $\mathcal{T}$ where the open sets are $\emptyset,X$ along with the sets of the form $\{x\in\lambda|x>\alpha\}$, then $\lambda$ is not $T\_{inj}$. Furthermore, if $\kappa$ is an uncountable regular cardinal, and
$f:\kappa\rightarrow\mathcal{T}$ is a function with $\alpha\in f(\alpha)$ for all $\alpha<\kappa$ must be constant on a stationary set by Fodor's Lemma. In this sense, the axiom $T\_{inj}$ fails badly for uncountable regular cardinals $\kappa$.


$\textbf{The relation to partial orders}$


Recall that the specialization ordering on a space $X$ is the preordering $\preceq$ where $x\preceq y$ iff $\overline{x}\subseteq\overline{y}$, and the specialization ordering $\preceq$ is a partial ordering if and only if $X$ is $T\_{0}$. 


Let $(X,\leq)$ be a partially ordered set. Then the smallest topology $\mathcal{T}$ with specialization ordering $\leq$ is the topology generated by the subbasis $\{(\downarrow x)^{c}|x\in X\}$. Then since $\mathcal{T}$ is the smallest topology with specialization ordering $\leq$, it is easy to see that $(X,\mathcal{T})$ is $T\_{inj}$ iff there is some $T\_{inj}$ topology $\mathcal{S}$ on $X$ with specialization ordering $\leq$. Therefore the $T\_{inj}$ axiom is very closely related to the underlying specialization ordering of a topological space so it should be thought of in terms of partial orders as well as topological spaces.


