Definition of open set with epsilon

Question

I'm having some trouble in understanding the definition of an open set.

Fix a set $X \subset R^N $. A set $A \subset X$ is open (relative to X) if for every $ x \in A$ there is an $\epsilon >0$ such that $|x'-x|<\epsilon$ and $x'\in X$ implies $x'\in A$

I don't get it. How could there be $|x'-x|$ such as there is no number $\epsilon$ greater as $|x'-x|$? Could somebody please give me an example of such situation?

Answer 1

Example: The set $A=(0,1)$ is open in $X=\mathbb{R}$.

Indeed, for every $x\in(0,1)$ there is an $\varepsilon = \min\{\frac{1-x}{2},\frac{x}{2}\}$ such that if $x'\in\mathbb{R}$ satisfies $|x'-x|<\varepsilon$ then

$$|x'-x|<\frac{x}{2}\Rightarrow x'>\frac{x}{2}>0$$ and $$|x'-x|<\frac{1-x}{2}\Rightarrow x'-x<\frac{1-x}{2}\Rightarrow x'<\frac{1+x}{2}<1$$

Which means that $x'\in(0,1)$. Therefore $A$ is open in $\mathbb{R}$.

Geometrically you show that for every point $x\in(0,1)$ you can find a small interval $(x-\varepsilon,x+\varepsilon)$ which is contained in $(0,1)$.

If you're still confused try to show that $A=(0,1]$ is not open (pick $x=1$ and show there is no $\varepsilon>0$ with the given property).

Answer 2

Of course that for each $x$ and each $x'$ in $\mathbb{R}^n$ there is some $\varepsilon$ greater than $\lvert x'-x\rvert$. You are misunderstanding the definition of open set. Let $A$ be a subset of $X$ (which, in turn, is a subset of $\mathbb{R}^n$). We say that $A$ is open (in $X$) if for every $x\in A$, there is some $\varepsilon>0$ such that whenever $x'\in X$ is such that $\lvert x'-x\rvert<\varepsilon$, then $x'\in A$ too.

For intance, if $X=\mathbb{R}^2$ and $A=\{(x,y)\in\mathbb{R}^2\,|\,x>0\}$, then $A$ is open in $X$ because, if $(x,y)\in A$ (that is, if $x>0$), if you take $\varepsilon=x$ (it makes sense, since $x>0$), then$$\bigl\lVert(x',y')-(x,y)\bigr\rVert<x\implies\lvert x'-x\rvert<x\implies x'>0\implies(x',y')\in A.$$

Answer 3

You may also put it this way. Denote $N_{\epsilon}(x)=\{x'\in\mathbb R^n\mid |x'-x|<\epsilon\}$. Following the given definition, $A$ is open in $X$, iff for all $x\in A$ there's a positive $\epsilon$ such that $$N_{\epsilon}(x)\cap X\subset A.$$

Example: Let $n=1$, $X=[0,1]$ and $A=(1/2,1]$. Then $A$ obviously is open in $X$.

PS: Feel free to compare this definition with the definition of an open set in $\mathbb R^n$: here a set $A$ is open iff for all $x\in A$ there's a positive $\epsilon$ such that $$N_{\epsilon}(x)\subset A.$$ In this case $X=\mathbb R^n$.

Answer 4

The condition says that for each $x \in A$ there is some $\varepsilon>0$ (which can ,and usually does, depend on $x$) such that if we ever have any point $x' \in X$ that has distance to $x$ that is $< \varepsilon$ we are guaranteed to have $x' \in A$ as well. It could be that there are no such points in $X$ (except $x$ itself which has distance $0$ to itself!) and then the condition is vacuously fulfilled. An example of this is $X=[0,1] \cup \{2\}$ in $\mathbb{R}$, where we can take $\varepsilon =1$ (or smaller) for $A=\{2\}$ and $x=2$. So $\{2\}$ is then relatively open in $X$.