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Here a problem in topology. Let $I= [0,1]\subset\Bbb{R}$ and $X=I^2$. Also let $T_1$ be the order topology on $X$ using the dictionary order, and $T_2$ be the subspace topology that $X$ inherits from $Y=\Bbb{R}^2$ using the dictionary topology. Is $T_1=T_2$?

I think $T_1=T_2$ because they are both the dictionary topology that is homeomorphic to the product of discrete topology in one dimension and the standard topology in another. But according to this post, $T_2$ is not compact and connected, while $T_1$ is. So $T_1\subset T_2$. I wonder which is the correct answer. Thanks.

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  • $\begingroup$ Can you clarify what you mean by dictionary topology? The definitions seem confusing to me. Are you taking in both cases the order topology with respect to the lexicographical order? $\endgroup$ Commented 12 hours ago
  • $\begingroup$ The order topology uses the lexicographical order. $\endgroup$ Commented 11 hours ago

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The connected components they mention in the post witnessing that $T_2$ is not connected are examples of sets that are open in $T_2$ but not in $T_1.$ For instance, the interval (in terms of the dictionary order) $[(0,0), (0,1)].$ This is the restriction of, say, the open interval $((0,-1), (0,2)),$ so it's open in $T_2.$ But it's not open in $T_1,$ since $(0,1)$ is a limit point of its complement.

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    $\begingroup$ @JohnLee the interval in the example is clopen. $\endgroup$ Commented 11 hours ago
  • $\begingroup$ @JohnLee More specifically, $T_2$ is a disjoint union of $[0,1]$ copies of $[0,1]$ (in the standard topology), whereas $T_1$ has the intervals in a dense order. If the order were discrete rather than dense then it would amount to the same thing... e.g. the order topology on $\mathbb N\times [0,1]$ is the same thing as the subspace topology when considering them as a suborder/subspace of $\mathbb R^2$ in the dictionary order. (Though the same wouldn't be true for $\mathbb N \times [0,1)$, which would be connected in the order topology, as it's just the half-real line $[0,\infty)$ .) $\endgroup$ Commented 10 hours ago
  • $\begingroup$ Yes, I understand the interval is clopen in $T_2$ and so $T_2$ is not connected. But $T_1$ is also the dictionary topology. Why the same interval is not clopen in $T_1$? $\endgroup$ Commented 9 hours ago
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    $\begingroup$ @JohnLee $T_1$ is the dictionary order topology. $T_2$ is not. A subspace of an order topology is not necessarily the order topology of the suborder (e.g. consider $[0,1)\cup [2,3)$ as a suborder of $\mathbb R$ and a subspace of $\mathbb R$... it is connected in the former, but not the latter). As for why the interval is not clopen in $T_1$, there is no least element greater than $(0,1)$, so $(0,1)$ is a boundary point of $[(0,0), (0,1)]$ and it is not open. (In fact, $T_1$ is connected, though it is not path-connected.) $\endgroup$ Commented 8 hours ago
  • $\begingroup$ @ spaceisdarkgreen, thanks. $T_2$ can inherit from any part of $\Bbb{R}$. So it contains clopen sets and can be unbounded (not compact). $T_1$ is $[0,1]^2$ and is bounded. So it does not have clopen sets and is compact. But in order to show, $T_1\neq T_2$, this is not enough. Can you discuss more on basis of these two topologies? If we can show that the basis of $T_2$ properly includes $T_1$, then it is done. $\endgroup$ Commented 8 hours ago

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