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On Proof Wiki, the definition of an open set is stated as

Let $(M,d)$ be a metric space and let $U\subset M$, then $U$ is open iff for all $y\in U$, there exists $\epsilon \in \mathbb{R}_{>0}$ such that $B_M(y;\epsilon) \subset U$

Then there's a remark that states

It is important to note that, in general, the values of $\epsilon$ depend on $y$. That is, it is not required that $\exists \epsilon \in \mathbb{R}_{>0}$ s.t. $\forall y \in U$ s.t. $B_M(y;\epsilon) \subset U$

I don't understand quite understand the remark, could someone please clearly explain this and perhaps provide an example? I believe I understand the definition, but the remark is throwing me off.

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4 Answers 4

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As an example consider for instance $\mathbb R^2$ with its usual (Euclidean) metric structure (so distance look like what we are used to in the plane). Now consider the set $\{x\in \mathbb R^2\mid d(x,(0,0))<1\}$. This is the interior of the circle with centre the origin and unit radius. Now, this set is open, since for any point $p$ in this set, there is a radius $\epsilon>0$ such that a the ball with centre $p$ and radius $\epsilon$ is entirely contained in the set. But, the radius $\epsilon$ depends on the position of the point $p$. When $p$ is very close to the circumference of the circle the value of $\epsilon$ must be chosen to be very small.

It should be noted that it is actually extremely rare that in a metric space a set will be 'uniformly' open, in the sense that there will be a single $\epsilon$ that fits for all points in the set. As an extra exercise, try to find examples of such situations.

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The subtlety is in $\exists \epsilon, \forall y$ and $\forall y,\exists \epsilon $. Do you know uniform continuity and continuity? You must be able to note the difference.

It should be clear why $\forall y,\exists \epsilon $ works. The point is same epsilon may not work for every y. Consider the interval $(1,2)$ for a point $1.2$, $\epsilon=0.1$ works, for $1.5$, $\epsilon=0.4$ works, The point is there need not be same epsilon which works for all.

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4  
Or to put the difference between $\forall\exists$ and $\exists\forall$ into a joke: "Every twenty minutes, a man gets run over by a car" - "Oh, that poor fellow" –  Hagen von Eitzen May 18 '13 at 22:53

The definition of open sets in terms of a metric states that for each point in an open set there'll be some open ball of radius $\epsilon > 0$ such that the ball is totally contained in the set. In other words, if $(M,d)$ is a metric space, a subset $U\subset M$ is open if for every $p \in M$ there's some $\epsilon >0$ with the property that $B(x; \epsilon)\subset U$. The remark says that the $\epsilon$ possibly depends on the point, so that you don't need to present some $\epsilon$ that works for every single point of $U$.

Since this can sound kind of abstract, I think an example in the plane can help a little. Consider $\mathbb{R}^2$ the euclidean plane with the usual euclidean metric. Given some subset $U\subset \mathbb{R}^2$ this set will be open if for every of it's points there's some open ball (which you can see for yourself in the plane will be discs without border). If the set $U$ is open, we can surround each point $p\in U$ with one of these balls so that all points sufficiently near $p$ will yet be in $U$. If you think for a while, points far from the border of the set $U$ can be surrounded by big discs, so that $\epsilon > 0$ is greater, and points nearer the border will need small $\epsilon$ to grant the disc still sits inside $U$.

Now, this is a good way to intuit that the $\epsilon$ depends on the point. I hope this helps you out. Good luck!

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The statement is like uniform continuity (in terms of structure). All its saying is if you get really close to the "border" of an open set, you don't have mush wiggle room to stay in it ($\epsilon$ has to be small to stay in the set), versus if you were in the "middle" of it.

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