Let $I = [0, 1]$. Compare the product topology on$ I × I $, the dictionary order topology on $I × I$ , and the topology$ I × I $inherits as a subspace of $R×R $in the dictionary order topology..
As i got the answer but did not understand in my head...https://dbfin.com/topology/munkres/chapter-2/section-16-the-subspace-topology/problem-10-solution
Pliz help me....
Thanks in advance
In the product topology on $I\times I$, sets of the form $U \times V$ are open, where $U$ and $V$ are open in $I$. $I = [0,1]$ is an open subset of itself, and $(0.5, 1]$ is the intersection of $(0.5, 2)$ with $I$, and is therefore an open subset of $I$. So their product $[0,1] \times (0.5,1]$ is an open subset of $I \times I$. It includes the point $\langle 0,1\rangle$ (where I am using the $\langle \, , \,\rangle$ notation for points, since this one of the rare times where using $(,)$ for both points and intervals really is confusing). So it is an open neighborhood of that point.
But if we look at the same subset in the dictionary order topology, not only is it not open, it contains no neighborhood of $\langle 0,1\rangle$. Since $0$ is the least element of $I$, in the dictionary order, the only points less than it are those with $0$ as first coordinate: $$\langle x,y\rangle \prec \langle 0,1\rangle \iff x = 0, y < 1$$ So an interval (basic open set) about $\langle 0,1\rangle$ will look like this: $$(\langle 0,a\rangle, \langle b,c\rangle) = \{\langle x, y\rangle \mid x < b \text{ or }(x = b \text{ and }y < c)\}$$ with $a < 1$ and $b > 0$. Note that for any $0 < u < b$, the entire set $\{u\} \times I$ is contained in this interval, but is not a subset of $[0,1] \times (0.5,1]$ Since this true for any open interval about $\langle 0,1\rangle, [0,1] \times (0.5,1]$ contains no dictionary-order neighborhoods of it.
In the other direction, if we choose $c = 0$, the interval $$(\langle 0,a\rangle, \langle b,0\rangle) = \{0\}\times (a,1] \cup (0,b) \times [0,1]$$ is open in the dictionary order, but does not contain points in $[0,\epsilon) \times \{1\}$, like every neighborhood of $\langle 0,1\rangle$ does in the product topology for sufficiently small $\epsilon$.