# Intervals that are open sets in $Y = (2,7]$ as a topological subspace of $\mathbb{R}$ with the standard topology

From Morris's Topology Without Tears, page 86,

Let $Y$ be a non-empty subset of a topological space $(X,T)$. The collection $T_y={O\cap Y : O \in T}$ of subsets of $Y$ is a topology on $Y$ called the subspace topology...The topological space $(Y,T_y)$ is said to be a subspace of $(X,T)$

So, if $Y=(2,7]$ and let $\mathbb{R}$ with the standard topology, the following intervals represent open sets in the subspace topology on Y: $(2,7)$,$(3,5)$, and $(6,7]$.

Alternatively, $(3,5]$ and $(1,6)$ are not open sets in the subsuspace topology on $Y$.

Is my understanding correct?

• $(2,7)$ is open because $(2,7)=(2,7)\cap Y$, and $(2,7)$ is an open set in $\mathbb R$.
• $(6,7]$ is open because $(6,7] = (6,8)\cap Y$ and $(6,8)$ is open in $\mathbb R$.
Also, $(3,5]$ is not open in $Y$ because there does not exist any open set $O$ such that $O\cap Y=(3,5]$. This is because for any open set $O$ in the standard topology over $\mathbb R$, if $5$ is an element of $O$, there exists some $0<\epsilon<1$ such that $(5-\epsilon, 5+\epsilon)\subseteq O$. However, this also means that $x_\epsilon = 5+\frac\epsilon2$ is an element of $O$, and (as $x_\epsilon$ is also it is also an element of $Y$), $x_\epsilon$ is also an element of $O\cap Y$. This means $O\cap Y\neq (3,5]$.
$(1,6)$ is not even a subset of $Y$ and so cannot be open in $Y$.