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So I've been working with the concepts of open and closed sets wrt a metric space $(X,d)$ and something thats confusing me a little is open balls. specifically given an epsilon ball $B_e:=\{y \in \Bbb R | d(x,y)<e\}$ then

i) when considering whether or not a set is open can we decide to make the epsilon arbitrarily large, as so long as the distance is less than any real number it is still open ?

ii) in the actual definition of an open set , namely , $A \subset X $is open if $ \forall x \in A \exists e>0 s.t. B_e(x) \subset A $ do we consider the epsilon to get smaller and smaller the closer to the boundary we get ?

iii) does this make the boundary of an open set a limit point ?

excalibirr
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2 Answers2

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i) A set $S$ is open, if and only if, for all its elements $x$ there is a $B_\epsilon(x) \subset S$. In more wordier version. A set S is is open if and only if for each one its elements there is at least an open ball, centered in that element which is a subset of S.

ii) The epsilon you need is any that works. If you found one that worked, for that particular element, and you can find one for each element, then your set is open.

iii) boundary of an open set

I don't know if this would help but here you have an example.

Consider the set of real numbers, and the classical Euclid metric. Lets check if the next Sets are open:

  1. [0,1] is not an open set in ($\mathbb{R},d$). One of the points of the given set is 1. You can not find any $B_\epsilon(1) \subset [0,1]$. Get the ball centered in 1, can you find any $\epsilon > 0$, that assures that any Real between $[1-\epsilon, 1+\epsilon]$ is completely inside $[0,1]$? The answer is no you would need $\epsilon=0$.

  2. (0,1) is an open set. you can proceed in a similar way that before. imagine you get any $0<x_0<1$, you can always get the ball $B_{\min(1-x_0,x_0)}$ and assure that all its elements are inside interval (0,1).

  3. $\mathbb{Q}$ is not open in the reals with the euclid distance, because no matter which ball you get, centered on any rational number it will always have a real number inside.

abr
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  • Something very important, just a trick, never ever consider that the opposite of being open is being closed, because it is not. – abr Jan 09 '18 at 21:14
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I hope I understood your question correctly, but here:

For i), $B_{\epsilon}(x)$ is an open set regardless of how large or small $\epsilon$ is. The only restriction is $\epsilon > 0$.

For ii), suppose $y$ is a point of the boundary of $A$. Then for a point $x \in A$, to have $B_{\epsilon}(x) \subset A$, we must have $\epsilon < d(x,y)$. See if you can prove this.

And for iii), the boundary of an open set is a $\textit{set}$, not a limit point. The boundary consists of points that may or may not be limit points. For example, if $A = (0,1] \cup \{2\}$, then the boundary of $A$ is $\{0,1,2\}$. $0$ is a limit point of $A$ that is not contained in $A$. $1$ is a limit point of $A$ that is contained in $A$. $2$ is not a limit point of $A$.

Mark
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  • you answered part i) and ii) very succinctly however for i) I'm afraid that's not quite what I meant. I meant given some set $A \subset X$ if we aim to prove that A is open we would show that it does not contain @Mark it's endpoints by constructing an epsilon ball , so say we had A is the set (0,100) and X=$\Bbb R$ then could we say epsilon=50 ( and I know its usually a placeholder, and we wouldnt actually call it any number) as theres no point that could have a radius from the endpoints which was not less than that ? – excalibirr Jan 09 '18 at 20:57
  • With the example $A = (0,100)$ and $X = \mathbb{R}$, you have to show that for every point $x \in A$, there is some $\epsilon > 0$ ($\epsilon$ is allowed to depend on the point $x$) such that $B_{\epsilon}(x) \subset A$. $\epsilon = 50$ works only for the point $x = 50$. In general, you could say "let $\epsilon_{x} = \min{x, 100-x}$." I wrote $\epsilon_{x}$ to emphasize that $\epsilon$ can depend on $x$. This way, no matter what $x$ you choose in $A$, we have $B_{\epsilon_{x}}(x) \subset A$. – Mark Jan 09 '18 at 21:03