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Consider the subset $ \ Y=[0,1) \cup \{2 \} \ \subset \mathbb{R} $.

Show that $ \ \{2 \} \ $ is open in subspace topology in Y but not open in order topology in Y.

Answer:

$ (\frac{3}{2}, \frac{5}{2} ) \cap Y=\{2\} \ \Rightarrow \{2 \} \ $ is open in the subspace topology in $ \ Y \ $.

But I do not know what will be the form of any basic open set containing $ \ 2 \ $ in the order topology in $ \ Y \ $

Is it of the form $ \ \{x | x \in Y , \ a<x \leq 2 \} \ $ ?

If then , how?

kindly help me with this problem

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  • $\begingroup$ Are you sure it's not $Y = [0,1) \cup \{2\}$? $\endgroup$ – Daniel Fischer Apr 14 '18 at 15:34
  • $\begingroup$ yes it is $ \ [0,1) \{2 \} \ $ . $\endgroup$ – M. A. SARKAR Apr 14 '18 at 15:44
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With respect to the order topology, if $A$ is an open subset of $Y$ and $2\in A$, then $A\supset(a,1]\cup\{2\}$, for some $a\in[0,1]$. Therefore, $\{2\}$ is not an open set with respect to that topology.

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    $\begingroup$ Because, by definition a subbase of the order topology are the sets of one of these type: $\{x\in Y\,|\,x>a\}$ and $\{x\in Y\,|\,x<a\}$ (for some $a\in Y$). But the sets of the second of these types don't contain $2$. So, any open set must contain a set of the first type. $\endgroup$ – José Carlos Santos Apr 14 '18 at 15:27
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    $\begingroup$ But $(1,+\infty) = \{ x \in Y : x > 1\} = \{2\}$ looks open to me. $\endgroup$ – Daniel Fischer Apr 14 '18 at 15:33
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    $\begingroup$ @DanielFischer Oops! You're right. My argument would work if $Y=[0,1)\cup\{2\}$, but not for the $Y$ provided by the OP. I shall delete my answer. $\endgroup$ – José Carlos Santos Apr 14 '18 at 15:35
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    $\begingroup$ Perhaps wait for the OP to clarify and edit appropriately. $\endgroup$ – Daniel Fischer Apr 14 '18 at 15:36
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    $\begingroup$ @DanielFischer Thank you. I will wait. $\endgroup$ – José Carlos Santos Apr 14 '18 at 15:39
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The basic elements of the order topology of a given set X, are the collection of intervals : $$(a,b) := \{c \in X | a<c<b\}$$ $$[a_0,b) := \{c \in X | a_0\leq c<b, \text{where } a_0 \text{ is the smallest element of } X\}$$ $$(a,b_0] := \{c \in X | a_0<c\leq b, \text{where } b_0 \text{ is the biggest element of } X\}$$

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