# How are open subsets and relatively open subsets different from each other?

To my understanding, if we have a set $$U$$ and a $$A\subset U$$ then $$A$$ is relatively open to $$U$$ if $$\exists$$ and open set $$\Omega\subset U$$ s.t $$A=\Omega \cap U$$. But also a subset of $$U$$ is open $$\iff$$ it is relatively open so would it isn't wrong to say that $$\Omega = A$$ so why do we need to define the idea of openness relative to a set?

• $U$ is open if it is relatively open to what set? The thing about relative openness is says something about how we are considering the set in question. – Aaron Zolotor Oct 21 '18 at 17:34
• I missed out part of the question! Thanks for pointing that out! – BigWig Oct 21 '18 at 17:36

Take for example $$\mathbb{R}$$ with the standard topology. $$A = [0, 1)$$ is not an open, but if you take $$U = [0, 2]$$, you see that $$A$$ is relatively open with respect to $$U$$, because $$[0, 1) = [0, 2] \cap (-1, 1)$$, which is the intersection of $$U$$ and an open subset of $$\mathbb{R}$$.
• Oh!! so $\Omega$ itself doesn't need to be a subset of $U$? Would it also be that my assumption $A$ is open iff $A$ is relatively open in $U$ is false then? – BigWig Oct 21 '18 at 17:41
• No, $\Omega$ is a subset of $\mathbb{R}$, in this case. More generally, if you had $U \subseteq V$ (both topological spaces), then $\Omega$ is a subset of $V$. This is known as induced or subspace topology. Yes, it's false, the example I showed above gives you a set not open in $\mathbb{R}$, but that's open in the subspace topology of $[0, 2]. – user512979 Oct 21 '18 at 17:47 Consider $$\mathbb R$$ with usual topology and $$(0,1]$$ with subspace topology wrt $$\mathbb R$$ Note that $$(1/2,1]$$ is open in $$(0,1]$$ but not open in $$\mathbb R$$ i.e. $$(1/2,1]$$ is relatively open in $$(0,1]$$ • So would it be that in your example$\Omega\$ = A = (1/2, 1] in this case? – BigWig Oct 21 '18 at 17:40