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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?

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  • $\begingroup$ What is objectionable about this post, I wonder? $\endgroup$ May 10, 2015 at 9:59
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    $\begingroup$ 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? $\endgroup$ May 11, 2015 at 0:43
  • $\begingroup$ @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. $\endgroup$ May 11, 2015 at 13:42
  • $\begingroup$ @Brian M. Scott, can you please check my work too? Your feedback is always so illuminating! $\endgroup$ May 11, 2015 at 13:43

1 Answer 1

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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.)

The other two properties to check follow from these.

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  • $\begingroup$ He said $\mathbb{R}^{2}$ in the lexicographical order topology. A vertical line is an open set in this topology. $\endgroup$
    – JKEG
    May 11, 2015 at 0:52
  • $\begingroup$ 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. $\endgroup$ May 11, 2015 at 1:00

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