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

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.

$(\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

• Are you sure it's not $Y = [0,1) \cup \{2\}$? – Daniel Fischer Apr 14 '18 at 15:34
• yes it is $\ [0,1) \{2 \} \$ . – M. A. SARKAR Apr 14 '18 at 15:44

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.
• 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. – José Carlos Santos Apr 14 '18 at 15:27
• But $(1,+\infty) = \{ x \in Y : x > 1\} = \{2\}$ looks open to me. – Daniel Fischer Apr 14 '18 at 15:33
• @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. – José Carlos Santos Apr 14 '18 at 15:35
The basic elements of the order topology of a given set X, are the collection of intervals : $$(a,b) := \{c \in X | a<c<b\}$$ $$[a_0,b) := \{c \in X | a_0\leq c<b, \text{where } a_0 \text{ is the smallest element of } X\}$$ $$(a,b_0] := \{c \in X | a_0<c\leq b, \text{where } b_0 \text{ is the biggest element of } X\}$$