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The definition of an open set in a Euclidean space $\mathbb{R}^n$ as stated in Wikipedia is:

a subset $U \subset \mathbb{R}^n$ is open if every point in $U$ is the center of an open ball contained in $U$

What I want to know is why specifically an open ball is required. Why not a closed ball? Because the way I see it - if you can show that a point in the set is the center of an open ball, you could always define a smaller closed ball around it. So why not take closed balls instead?

Also, from here: Can an open ball have just one point. As per my understanding it cannot. Please clarify. , I understand that open balls containing just one point are also possible. So, possibly, even for a closed set, we could define points on the boundary as being the center of an open ball. So (to avoid this), should the definition also state explicitly that the open balls should have more than one point. OR is the structure of the Euclidean space such that open balls containing one point do not exist.

K.defaoite
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karun mathews
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3 Answers3

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EDIT: My initial answer was incorrect; the two characterizations are indeed equivalent.

See the accepted answer on this post: Is this statement true: a set is open if every point has a closed ball contained inside of the set.

I.e., you can take either open or closed balls.

masiewpao
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    Actually, in the example you use to show that closed balls can't be used to characterize open sets - can't we use a smaller closed ball instead of $B_d (2;1)$ like $B_d (2;0.5)$. Then wouldn't such a definition work? – karun mathews Jul 03 '20 at 15:21
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    Because we just require one closed ball lying completely inside the set $O$ right? – karun mathews Jul 03 '20 at 15:22
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    @karunmathews no you are absolutely correct, and what I said is wrong; it doesn't answer your question. I think I can maybe explain it in terms of the properties, let me edit my answer. – masiewpao Jul 03 '20 at 15:45
  • ok thanks for the help @masiewpao – karun mathews Jul 03 '20 at 15:51
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If you allow closed balls, then you also allow singleton sets: a closed ball of radius zero around a point $x\in\mathbb{R}^{n}$ looks like $$\{y\in\mathbb{R}^{n}:\lVert y-x\rVert\leq0\} = \{x\}.$$ We don't want singletons to be open sets (if all singletons are open, then every set is open, because a union of open sets should be open as well). It's cleaner to just focus on open balls rather than have to qualify which closed balls are allowed.

Will R
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To be a definition you need CONDITIONS $\iff $ DEFINITION

If you have DEFINITION $\implies$ OUTCOME then you don't necessarily have OUTCOME $\implies$ RESULT.

So you can't use OUTCOME as a definition.

We have OPEN SET $\iff$ every point has an open ball around it $\implies$ every point has a closed ball around it

but we don't have Every point has a closed ball around it $\implies$ every point has an open ball around it.

(Note singletons are closed balls so every set has the property. If you modify the property with non-singleton closed ball or closed ball with positive radius... I think then either would be acceptable but... why confuse things?)

fleablood
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