# Can we characterize those topological spaces (with more than one point) in which every two point subset is connected but not path-connected?

Can we characterize those topological spaces (with more than one point) in which every two point subset is connected but not path-connected ? The only progress I've made is that such a space cannot be $T_1$ . Please help . Thanks in advance

• Here is a complete list of the spaces with that property: $\varnothing$. Every connected two-point space is path-connected. – Daniel Fischer Oct 3 '15 at 11:53

Note: An earlier edition of this answer was rooted in a false belief that the space $\{a, b\}$ with topology $\{\emptyset, \{a\}, \{a, b\}\}$ is not path connected. But actually the path $$\gamma(t) = \begin{cases} a & \text{if } t \in [0,1) \\b & \text{if } t = 1 \end{cases}$$ is continuous and connects $a$ and $b$.

So the short answer to this question is given by our moderator in the comment already.

If you like it, the answer below is really answering the question:

What can you say about $X$, where all two points set $\{a, b\}$ of $X$ with topology $\{\emptyset, \{a\}, \{a, b\}\}$ or $\{\emptyset, \{b\}, \{a, b\}\}$?

Define a relation on $X$ by $x< y$ if and only if the topology on $\{x, y\}$ is given by $\{\emptyset, \{x\}, \{x, y\}\}$. It's easy to see that $x<y$ if and only if there is an open set $V$ so that $x\in V$ and $y\notin V$.

It is also easy to see that for all $x, y\in X$, either one of them holds: $$x< y, \ \ x =y,\ \ y<x.$$

Claim 1: $<$ is transitive: Let $x <y$ and $y<z$. Then there is open set $V_x$ of $X$ so that $x\in V_x$ and $y\notin V_x$. Let $U$ be an open set containing $y$. Then $x\in U$ (or $\{x, y\}$ would be given the discrete topology). In particular, we have $x<z$ (Indeed, $V_x \subset U$ as $p<y$ for all $p \in V_x$).

Thus we are able to give $X$ a strict total order.

Claim 2: Let $x\in X$. Then $U_x = \{y\in X: y<x\}$ is open.

To see this, let $y\in U_x$. Then there is $V_y$ so that $y\in V_y$ and $x\notin V_y$. But then $z<x$ for all $V_y$ by definition. Thus $V_y\subset U_x$. This shows $U_x$ is open.

Thus, we know that the topology of $X$ must be finer than the left order topology.

Claim 3: Let $U\subset X$ be nonempty open, $U\neq X$. Then there is $y_2$ so that $$U \subset U_{y_2}.$$

To see this, let $y_2\notin U$. Then if $y\in U$, then $y < y_2$. Thus $U \subset U_{y_2}$. (Note that $U$ might not contains any $U_x$)

I am not sure what's more can be said about this. Indeed, the topology could be strictly bigger (than the left open topology). For example, take $X = \mathbb R$ with the usual ordering and with the topology generated by

$$W_x = \{y\in \mathbb R: y\le x\}.$$

This topology on $\mathbb R$ satisfies your condition, and is not generated by the left order topology.

• But the Sierpiński space is path-connected. Every connected two-point space is path-connected. – Daniel Fischer Oct 3 '15 at 11:49
• @DanielFischer : You are absolutely correct. Should have checked that carefully. Thanks. – user99914 Oct 3 '15 at 11:55