# Prob. 3, Sec. 25 in Munkres' TOPOLOGY, 2nd ed: Is $I \times I$ path connected or locally path connected in the subspace topology?

Let $$I \ = \ [0,1] \ = \ \{\ x \ \in \mathbb{R} \ \colon \ 0 \leq x \leq 1 \ \}.$$

In the subspace topology that $I \times I$ inherits from the dictionary order topology on $\mathbb{R} \times \mathbb{R}$ is $I \times I$

(i) loaclly connected?

(ii) locally path connected?

(iii) connected?

(iv) path connected?

I know that if $I \times I$ is (locally) path connected, then it would also be (respectively, locally) connected. So first we should check (local) path-connectedness.

My effort:

I think $I \times I$ in the subspace topology is not connected:

Let $A \colon= \{0\} \times I$, and let $B \colon= (0, 1] \times I$. Then $A \cap B = \emptyset$, and $A \cup B = I$. Moreover, both $A$ and $B$ are non-empty.

Now $$A = \{0 \} \times I = (I \times I ) \ \cap \ (\ 0 \times -1, \ 0 \times 2 \ ),$$ where $$(\ 0 \times -1, \ 0 \times 2 \ ) \ \colon= \ \{ \ x \times y \in \mathbb{R} \times \mathbb{R} \ \colon \ 0 \times -1 \ < \ x \times y \ < \ 0 \times 2 \ \},$$ which is open in the dictionary order topology on $\mathbb{R} \times \mathbb{R}$. So $A$ is open in the subspace topology on $I \times I$. Am I right?

Now
$$B = (0, 1] \times I = (I \times I) \cap ( 0 \times 2, 1 \times 2 ),$$ where $$( 0 \times 2, 1 \times 2 ) \ \colon= \ \{ \ x \times y \ \in \mathbb{R} \times \mathbb{R} \ \colon \ 0 \times 2 < x \times y < 1 \times 2 \ \},$$ which is an open set in the dictionary order topology on $\mathbb{R} \times \mathbb{R}$, so that $B$ is open in the subspace topology on $I \times I$. Am I right?

Thus, $A$ and $B$ form a separation of $I \times I$.

Hence $I \times I$ in the subspace topology is not connected and therefore is not path-connected either.

Have I reached a correct conclusion?

Now for local connectedness.

Let $x \times y \in I \times I$, and let $U$ be an open set in the subspace topology on $I \times I$ such that $x\times y \in U$. Let's even particularise $U$ to be a basis element for the subspace topology on $I \times I$; this leads to no loss of generality. Then $$U = (I \times I ) \cap (a \times b, a \times c).$$ for $a, b, c, in \mathbb{R}$. Sets of the form $(a \times b, a \times c)$ form a basis for the dictionary order topology $\mathbb{R} \times \mathbb{R}$. Am I right?

Thus $x = a$ and $b < y < c$; in fact, $\max(0, b) \leq y \leq \min(1, c)$. Since $U$ is non-empty, we must have $a \in I$ and also $(b, c) \cap I \neq \emptyset$; in fact, even $\max(0, b) < \min (1, c)$.

Thus, we can write $U$ as $$U = \{a \} \times [ \max (0, b) , \min(1, c) ].$$ So $U$, being homeomorphic with a closed interval on the real line, is connected.

Hence $I \times I$ in the subspace topology is locally connected also.

Is this the correct conclusion? Have I managed to get all the steps and statements right?

Now for local path-connectedness:

I guess $I \times I$ is locally path connected also.

Suppose that $x \times y \in I \times I$, and suppose that $U$ is a basis element containing $x \times y$, as before. Then, for some $a, b, c \in \mathbb{R}$, we have $$U = \{a \} \times [ \max(0, b), \min(1, c) ],$$ which, being homeomorphic with a closed interval on the real line, is path-connected.

Is the reasoning correct? Is this conclusion correct?

• This post has received downvotes because it does not show enough of your attempts to solve the problem. You know that you need to check path connectedness: so what did you try? How did it work out? May 11 '15 at 0:43
• @Carl Mummert, please have a look at this post again. I've editted it to include my effort. Please check my work and then advise me on how good an attempt it is. May 11 '15 at 13:42
• @Brian M. Scott, can you please check my work too? Your feedback is always so illuminating! May 11 '15 at 13:43

It is not connected, the intersection of $I\times I$ with a vertical line, say $y=\frac{1}{2}$, is a nontrivial clopen.
To check it is locally path connected, show that intervals are path connected, and see that given any point $x\in I\times I$ and open set $U$ containing $x$, there is an interval containing $x$ and contained in $U$. (Or more formally, a set homeomorphic to an interval.)
• He said $\mathbb{R}^{2}$ in the lexicographical order topology. A vertical line is an open set in this topology.
• Thanks. I misread the entire problem! I thought it was asking about the irrational square as a subspace of $\mathbb{R}^2$ in the regular topology! I have no idea how I read that. May 11 '15 at 1:00