Are minimal neighborhoods in an Alexandrov topology path-connected?

Question

An Alexandrov topology, AKA an Alexandrov discrete topology, is a topology in which the arbitrary intersection of open sets is open. (As opposed to other typologies where only the finite intersection of open sets is open.) Now, in such a topology, for any element $x$, the intersection of all the open sets containing $x$ is open, so let us call it the minimal neighborhood of $x$.

My question is, is the minimal neighborhood of $x$ always path-connected for any point $x$? It's clearly connected, but a connected set need not be path-connected.

Answer 1

Yes. More generally, if $X$ is a topological space with a point $x\in X$ which is in every nonempty closed subset of $X$, then $X$ is contractible. The minimal open neighborhood $X$ of a point $x$ in an Alexandrov space has this property, because if $C\subseteq X$ were nonempty and closed and did not contain $x$, then $X\setminus C$ would be a smaller open neighborhood of $x$.

The proof is simple: define $H:X\times [0,1]\to X$ by $H(a,t)=a$ for all $a\in X$ and $t<1$ and $H(a,1)=x$ for all $a\in X$. This is continuous: if $C\subseteq X$ is closed and nonempty, then $x\in C$ and so $H^{-1}(C)=X\times \{1\}\cup C\times [0,1]$ is closed in $X\times [0,1]$. So, $H$ is a homotopy from the identity map on $X$ to the constant map with value $x$, so $X$ is contractible.

Explicitly, you get a path from any point $a\in X$ to $x$ by restricting $H$ to $\{a\}\times[0,1]$. So, the path takes value $a$ for all $t<1$ and value $x$ at $t=1$.

(A similar construction also shows that a topological space with a point which is in every nonempty open subset of $X$ is contractible, so that the closure of a point in any topological space is contractible.)

Answer 2

If $(X, \mathcal{T})$ is an Alexandrov space define for each $x \in X$ : $M_x = \bigcap \{O \in \mathcal{T}: x \in O\}$, which is the minimal open neighbourhood of $x$, as stated.

Now consider $Y= M_x$ for some point $x$. We want to see that $Y$ is path-connected, so it suffices to find a continuous path from $x$ to any $y \in Y$, where we can assume WLOG that $y \neq x$ (take a constant map otherwise).

Now consider the subspace $\{x,y\}$ of $Y$. There are but two options for its topology: consider $M_y$. As $y \in M_x$ and $M_x$ is open, $M_y \subseteq M_x$ so we either have $M_x = M_y$ (which happens when $x \in M_y$) or $M_y \subsetneq M_x$ (which happens if $x \notin M_y$). In the former case the subspace topology on $\{x,y\}$ is indiscrete. In the latter case we get a Sierpiński topology $\{\emptyset, \{y\}, \{x,y\}\}$ on $\{x,y\}$.

In both cases the "jump path" : $p: [0,1] \to \{x,y\}$ defined by $p(0) = x$, and $p(t) = y, t>0$ is easily seen to be a continuous path from $x$ to $y$, and as the continuity only depends on the image subspace, $p$ seen as a function from $[0,1]$ to $Y$ is still continuous and is the required path showing $Y$ path connected.

So the crux is really that $Y$ has no discrete two point subsets.