Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
$N_{\epsilon}$ is any (open) ball containing $x$ of radius $\epsilon$ with $x$ in the center. –  user59671 Feb 6 '13 at 17:25

2 Answers 2

up vote 6 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.

share|improve this answer
1  
@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 $X$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.