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I came across the following:

Definition 15. Let $X$ be a subset of $\mathbb{R}$. A subset $O \subset X$ is said to be open in $X$ (or relatively open in $X$) if for each $x \in O$, there exists $\epsilon = \epsilon(x) > 0$ such that $N_\epsilon (x) \cap X \subset O$.

What is $\epsilon$ and $N_\epsilon (x) $? Or more general, what are relatively open sets?

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$N_{\epsilon}$ is any (open) ball containing $x$ of radius $\epsilon$ with $x$ in the center. – user59671 Feb 6 '13 at 17:25
up vote 8 down vote accepted

Forget your definition above. The general notion is:

Let $X$ be a topological space, $A\subset X$ any subset. A set $U_A$ is relatively open in $A$ if there is an open set $U$ in $X$ such that $U_A=U\cap A$.

I think that in your definition $N_\epsilon(x)$ is meant to denote an open neighborhood of radius $\epsilon$ of $x$, ie $(x-\epsilon,\ x+\epsilon)$. As you can see, this would agree with the definition I gave you above.

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@Omar - The definition you gave is specific to metric spaces. Daniel's definition will work in any topological space and hence is of greater use. – Chris Leary Feb 6 '13 at 17:35
@Daniel Robert-Nicoud - Suppose I want to prove something, e.g. continuity, such that I would need to find some open set $U$. Would it be sufficient to show that there exists a relatively open set? – omar Feb 6 '13 at 19:00
@omar: What would you like to prove exactly? Give me a more specific question and I might (should) be able to help you. – Daniel Robert-Nicoud Feb 6 '13 at 19:07
@DanielRobert-Nicoud - In Urysohn's Metrization Theorem we have some space X and construct a function $F: X \rightarrow H$, where $H$ is the hilbert cube. The idea is then to show that $F$ is an embedding by proving $F$ to be one-to-one, continuous and open. When proving continuity let $W$ be open in $X$, then they show that $F(W)$ is the union of relatively open sets. So normally when proving continuity one would need to find an open $F(W)$, but here they suffice with the union of relatively open sets. – omar Feb 6 '13 at 20:18
@omar: Actually to do what you say is to prove that $F$ is an open map. To prove it is continuous you take any relatively open subset $U$ of its image and you have to show that its preimage $F^{-1}(U)$ is open. – Daniel Robert-Nicoud Feb 6 '13 at 20:28

Recall that generally, $O$ is open if for every $x\in O$ there exists some $\varepsilon>0$ such that $N_\varepsilon(x)\subseteq O$.

Being open relative to $X$ means that there is some open set $O'$ such that $O=O'\cap X$, and equivalently for every $x\in O$ there is some $\varepsilon>0$ such that $N_\varepsilon(x)\cap X\subseteq O$.

For example $O=\{0\}$ is not open in $\mathbb R$, but if we consider $X=\{0\}$ then for $\varepsilon=1$ we have that $N_\varepsilon(0)\cap X\subseteq O$, and therefore it is open relative to $X$.

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