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Here is Example 3, Sec. 16 in the boom Topology by James R. Munkres, 2nd edition:

Let $I = [0, 1]$. The dictionary order on $I \times I$ is just the restriction to $I \times I$ of the dictionary order on the plane $\mathbb{R} \times \mathbb{R}$. However, the dictionary order topology on $I \times I$ is not the same as the subspace topology on $T \times I$ obtained from the dictionary order topology on $\mathbb{R} \times \mathbb{R}$! For example, the set $\left\{ 1/2 \right\} \times \left( 1/2, 1 \right]$ is open in $I \times I$ in the subspace topology, but not in the order topology, . . .

I think I've properly understood this example.

Now my question is, can we find a subset of $I \times I$ that is open in $I \times I$ in the dictionary order topology but not in the subspace topology?

My feeling is that there isn't any such set. Am I right? If so, then can we prove this rigorously?

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If $A\subseteq (X,<)$ is a subspace where $X$ has the order topology and $A$ has a subspace topology $\mathcal{T}_A$ induced from $\mathcal{T}_<$, and $<_A$ is the restricted order. Then $\mathcal{T}_{<_A} \subseteq \mathcal{T}_A$.

Proof: a subbase for $\mathcal{T}_{<_A}$ are sets of the form $(a_1, \rightarrow)_{<_A}$ and $(\leftarrow, a_2)_{<A}$, where $a_1, a_2 \in A$. But by definition these segments equal $(a_1, \rightarrow)_{<} \cap A$ and $(\leftarrow, a_2)_< \cap A$ and so are in $\mathcal{T}_A$.

The inclusion can be strict, (except for convex sets $A$ and open sets $A$, IIRC), but there cannot be a relative order open set that is not subspace open.

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