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Let X be a non-degenerate ordered set in the order topology. Let $a < b \in X$. Then the closed interval $[a,b]$ has a subspace topology induced by the order topology on X. But $[a,b]$ has the order induced by the order on X. So $[a,b]$ also has an order topology induced directly by the inherited order. Are these two topologies on $[a,b]$ the same?

This is what I am supposed to show or prove, and above this is what I did. Please note, $S_0$ and $S_1$ are the basis I declared, $T_0$ is the topology generated by a basis on X. $T_1$ and $T_2$ are subspace and order topologies obtained on X. Then, I tried 2 show that they are equal. Do I need to show them in this way:

$V = U K$ = $(-\infty, \infty)$, and W = intersection of $(-\infty, \infty)$ with $[a,b]$ = $(a,b)$ ??

Any help will be greatly appreciated.

I presented this in class, my prof told me to complete this since this is incomplete. He is saying that you need to prove the other way around showing that $T_1$ is a subset of $T_2$ and the opposite of that. Then, you can conclude that $T_1$ = $T_2$.

I can explain what I did, I took $B_0$ to be a basis for a topology on X, and $B_1$ to be a basis for the order topology so that I get the sub-basis of $B_0$ as $(-\infty, \infty)$ and sub-basis of $B_1$ as $[a,b]$. Then, I took $T_0$ to be the topology generated by a basis, $T_1$ to be the subspace topology and $T_2$ to be the order topology. Then, I let $W$ which is a subset of $[a,b]$ to be in $T_1$ or ($W \in T_1$), and went on to show that its in basis $B_1$ but I got lost after that.

Can someone help me to fix this?? Any help will be greatly appreciated.


1 Answer 1


I don't follow your proof very well, try to give it more structure and better notation?

Something more general is true (and useful to know):

Suppose $(X,\le)$ is a LOTS (linearly ordered space in the order topology) and $A \subseteq X$ is order convex, which means:

$$\forall x,y \in A \forall z \in X: (x \le z \le y) \to (z \in A)\tag{1}$$

(Note that $[a,b]$ is indeed a standard example of an order convex subset of $X$.) then the subspace topology $\mathcal{T}_A$ that $A$ gets from $X$ is the same as the order topology $\mathcal{T}_{\le_A}$ that $A$ gets from its inherited order $\le_A = \le \cap (A \times A)$.

We have two inclusions, one of which uses the order convexity:

Let $a \in A$, then $(\leftarrow, a)_{\le_A}= (\leftarrow,a)\cap A$ if we spell out the definitions, and this means that all subbasic elements of $A$ in the order topology are also open in $\mathcal{T}_A$, as intersection of order-open sets with $A$. Hence

$$\mathcal{T}_{\le_A} \subseteq \mathcal{T}_A$$

regardless of $A$, really.

To see the reverse, let $O \subseteq A$ be open in $\mathcal{T}_A$ so that $O = O' \cap A$ for some (order) open $O'\subseteq X$. We must show that $O$ is $\le_A$-open so let $a \in O$. In particular $a \in O'$ so by the definition of the standard base for the order topology there is some open interval $(c,d)$ containing $a$ such that $(c,d) \subseteq O'$ (or a set of the form $[a,d)$ if $a = \min(X)$ or $(c,a]$ if $a=\max(X)$, but we'll postpone those cases for now).

If now we have some $a_L \in A \cap [c,x)$ and some $a_R \in A \in (x,d]$ we see that $(a_L,a_R)_{\le_A} \subseteq A \cap O'=O$, so $a$ is an interior point of $O$ in $\mathcal{T}_{\le_A}$. This is the easiest case. If $a_L$ exists but no $a_R \in (a,d] \cap A$, this means there cannot be some $a' \in A$ with $a' > a$, or else $(1)$ would have forced $d \in A$, which is not the case, so we conclude $(a_L, \rightarrow)_{\le_A} \subseteq A \cap O' = O$ again allowing the conclusion that $a$ is an interior point of $O$ in $\mathcal{T}_{\le_A}$. The subcase of having no $a_L$ but some $a_R$ yields a similar conclusion using the $\le_A$-subbasic set $(\leftarrow,a_R)_{\le_A}$. The final case is that no $a_L$,$a_R$ at all exists, so that also no other elements of $A$ at all can exist (using the order convexity again) and $A$ is really singleton and so $O=\{a\}$ is trivially open in $\mathcal{T}_{\le_A}$.

If $a = \max(X)$, so $a=\max(A)$ too, we have $(c,a] \subseteq O'$ for some $c \in X$, and can distinguish two subcases: some $a_L \in [c,a) \cap A$, and we can use $(a_L,a]$ in $\mathcal{T}_{\le_A}$ or no such $a_L$ and thus $a=\min(A)$ by order convexity and $A=\{a\}$ again.

The case $a=\min(X)=\min(A)$ is similarly boring.

In all cases we have shown $O \in \mathcal{T}_{\le_A}$ and so

$$\mathcal{T}_A \subseteq \mathcal{T}_{\le_A}$$

finishing the proof. Such subcase distinctions are quite common in ordered spaces proofs, as the basis elements vary for minimal and maximal points.

As an example where we have a non-order convex set $A$ in $X$ where we still have this equality of topologies on $A$, consider $A = \mathbb{Z}$ in $\Bbb R$: both topologies yield the discrete topology on $A$.

To see that order convexity is important in general, we can consider in $\Bbb R$ the subspace $A=[0,1) \cup \{2\}$, where $\{2\}$ is open in the subspace topology on $A$ but not in the order topology on $A$ (see the special role of a max again). Or take $A=[0,1]\times [0,1]$ as a subset/subspace of $\mathbb{R}^2$ in the lexicographic order topology (an example that's treated in Munkres as well) where $A$ is disconnected and non-compact in the subspace topology it inherits but is compact and connected in the lexicographic order topology.


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