Jtdylktuy

Let $leq$ denote the dictionary order relation on $Itimes I$ determined by less than or equal to on $I=[a,b], a,bin mathbb R$, Let $mathscr T$ denote the order topology on $Itimes I$. Then $(Itimes I,mathscr T)$ is locally connected but not locally pathwise connected

My attempt:-

For any neighbourhood of $pin I times I$, I can find a connected neighbourhood marked in blue. $p$ was arbitrary, so $Itimes I$. locally connected.
Won't the marked blue neighbourhood be pathwise connected?
Still I don't get a suitable point to show that given $Itimes I$ under dictionary order is not locally pathwise connected. Please help me.

I think your diagram has mislead you. You've forgotten to think about what happens if your point is for example $(0.5,1)$. (Let's assume our interval is the unit interval) Then any basic neighborhood of your point in the order topology will have upper bound $(0.5+epsilon,y)$ for some $epsilon > 0$ and $0le y le 1$.

Thus you need to be a bit more careful for both local connectedness and disproving path connectedness. First, though, we'll need some facts.

Infima and suprema exist in $Itimes I$.

Proof

By symmetry it suffices to show that infima exist. If $Ssubset Itimes I$, then let $s_1=inf pi(S)$, and let $s_2 = inf {yin I : (s_1,y) in S }$. Then I claim $(s_1,s_2)$ is the infimum of $S$. If $(a,b)$ is a lower bound for $S$, then
since $pi$ is order preserving, $a$ is a lower bound for $pi(S)$. Hence $ale s_1$. If $a < s_1$, then $(a,b) < (s_1,s_2)$, and we are done. Otherwise if $a=s_1$, then $(a,b)le (s_1,y) $ for all $yin I$ such that $(s_1,y) in S$. Hence
$ble y$ for all those $y$, so $ble s_2$. Hence $(a,b) le (s_1,s_2)$. $blacksquare$

Next, (closed or open or mixed) intervals in $Itimes I$ are connected.

Proof

Note first as a convenience that if $U$ is an open subset of $Itimes I$, then for all $tin U$ there exist $r,sin U$ such that $r<t<s$ and $(r,s)subseteq U$. This is true since if $a<b$ in $Itimes I$, then there exists $x$ in $Itimes I$ with $a<x<b$ (take $x$ to be the usual average of $a$ and $b$). Thus if $tin U$, we can find $r',s'$ such that $r'<t<s'$ and $(r',s')subseteq U$. But then choose $r,s$ with $r'<r<t<s<s'$, and then we have $r,sin U$ and $(r,s)subseteq U$ as desired.

Suppose for contradiction that $U$ and $V$ are nonempty open sets disconnecting our interval. Then let $uin U$, $vin V$ since both are nonempty. WLOG we can assume $u < v$. Thus we can reduce to the case where $[u,v]$ is a closed interval and $uin U$, $vin V$. Now let $t = inf V$. Clearly we have $u le t le v$, so $tin [u,v]$. Now if $tin V$, then there exist $r,sin V$ such that $r<t<s$, but then $t=inf V le r < t$, which is a contradiction, and similarly if $tin U$, then there are $r,sin U$ such that $tin (r,s) subseteq U$. But we know that $[u,t)subseteq U$ since $t$ is the infimum of $V$, but then $[u,s)subseteq U$, so
in fact $s$ is a lower bound for $V$ larger than $t$, contradiction. Thus all intervals in $U$ are connected. $blacksquare$

Corollary: $Itimes I$ is locally connected. $blacksquare$

Not locally path connected

Actually, I can't come up with a disproof of local path connectedness, but here's my start on it. Edit I found that this is example 6 in section 24 of Munkres and found a copy of it here and I've filled in the missing section.

To show $Itimes I$ is not locally path connected, we first prove the intermediate value theorem.

Let $(Omega,le)$ have the order topology on it. Let $a,bin Omega$. Let $C$ be a connected topological space and $f:Cto Omega$ be continuous with $a$ and $b$ in the image of $C$. Then $[a,b]$ is in the image of $C$.

Proof

If $cin [a,b]$ but not in the image of $C$, then $f^{-1}(-infty,c)$ and $f^{-1}(c,infty)$ are disjoint open sets disconnecting $C$. Contradiction. $blacksquare$

Now to show that $Itimes I$ is not locally path connected, we'll prove that there cannot be any path from $(0,1)$ to $(epsilon,0)$ for all $epsilon$ and $y$.

Proof

By contradiction. Suppose $f$ is such a path.
For all $xin (0,epsilon)$, $f^{-1}({x}times (0,1))$ is a disjoint open subset of $[0,1]$. But then we have uncountably many (nonempty, by the intermediate value theorem above) disjoint open subsets of $Bbb{R}$, and if we choose a rational in each of them, then we'd have uncountably many rationals. Contradiction. $blacksquare$