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

Question

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

Answer 1

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.

Answer 2

The basic elements of the order topology of a given set X, are the collection of intervals : $$(a,b) := \{c \in X | a<c<b\}$$ $$[a_0,b) := \{c \in X | a_0\leq c<b, \text{where } a_0 \text{ is the smallest element of } X\}$$ $$(a,b_0] := \{c \in X | a_0<c\leq b, \text{where } b_0 \text{ is the biggest element of } X\}$$