I am trying this problem.

4-10(From (J.Lee) Intoduction to Topological manifold)

Let $S$ be the square $I\times I$ with the order topology generated by the dictionary order (see Problem 4-6).

(a) Show that $S$ has the least upper bound property.

(b) Show that $S$ is connected.

(c) Show that $S$ is locally connected, but not locally path-connected.

I have done for (a) and (b) and am stuck at $(c)$.

My trial is ....

Recall that the order topology generated by the dictionary order $\leq $ is the topology that generated by sets \begin{align*} &G_{(a,b)}=\{(x,y)\in S : (x,y) > (a,b) \}&L_{(a,b)}=\{(x,y) \in S: (x,y) < (a,b) \} \end{align*}

for each $(a,b)\in I\times I$.


Let $p\in S$ be given and let a neighborhood $U\subseteq S$ of $p$ be given. Note that $L_p\cap U_p=\{p\}\subseteq U$ and it is open since each of them are two of the generator of this topology and the union is a finite intersection of open sets. Note that singleton set is clearly connected since we cannot find any two disjoint nonempty set of it. Thus, $S$ is locally connected.\

However, if it is true, then it is also locally connected since singleton set is also path connected... which is inconsistent with the problem that I need to prove. I don't see where I am wrong.

And help would be appreciated.

  • $\begingroup$ edited the strict $<$ which are correct for the subbase. $\endgroup$ Oct 25, 2018 at 16:55
  • $\begingroup$ @HennoBrandsma Thanks, I misread the definition. $\endgroup$
    – Lev Bahn
    Oct 25, 2018 at 17:06

1 Answer 1


You gave the wrong subbase for the order topology. They should be the strict left and right sets, not with endpoints, or all order topologies would be discrete.

You can check that all open intervals and half-open intervals, which form a base for the order topology, are all connected (they are still complete and have no jumps or gaps, like $I \times I$, and thus are connected, see e.g. my answer. This shows local connectedness.

But at points of the form $(a,1), a < 1$ or $(a,0),a>0$ have open intervals as a local base which span more vertical lines, and one can show that such intervals are not path-connected. Look for a proof on this site that $I \times I$ in the order topology is not path-connected (or look it up online, there are many proofs of this online) and adapt it to neighbourhoods of the aforementioned points.

  • $\begingroup$ Where will I get the proof of (c)? $\endgroup$
    – Math geek
    Nov 14, 2018 at 15:41
  • $\begingroup$ Isn't it path connected?If I consider an open neighborhood of $x\in I \times I$. Can't I find path joining horizontal and vertical lines between any two points? $\endgroup$
    – Math geek
    Nov 14, 2018 at 15:44
  • $\begingroup$ @Mathgeek no, because a path must span uncountably many vertical stalks, and this makes the image non-separable. $I \times I$ is a compact first countable connected hereditarily normal space which is not separable. $\endgroup$ Nov 15, 2018 at 21:13

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