# Local connectedness and path-connectedness of a square $I\times I$.

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

(c)

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.

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

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.
• 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? Commented Nov 14, 2018 at 15:44
• @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. Commented Nov 15, 2018 at 21:13