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As far as I know, the standard definition of an open set is that the set $A$ is called open if $A \subseteq X$ for some set X and if $A \cap \partial A=\emptyset$ where $\partial A$ is the set of boundary points of A. In particular, I fail to see the motivation for the $A \cap \partial A=\emptyset$ part of this definition, why wouldn't replacing this with $\partial A=\emptyset$ be satisfactory?

Admittedly, experience is telling me that this is a case of "we define it this way because it is useful", but if that is the case then in what way is this useful? After all, as far as I can tell the alternative that I've proposed is very nearly equivalent to the standard definition. The only difference that comes immediately to my mind is that the standard definition would treat the case of $A=\emptyset$ differently. Is that an important difference? Or have I missed something else that is important?

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    $\begingroup$ Consider $\mathbb{R}$. The number $2$ is on the boundary of the interval $(-2,2)$, as is the number $-2$. The boundary here is not empty, right? $\endgroup$ – Matt Dec 6 '17 at 23:58
  • $\begingroup$ Excellent point. My proposed definition does not agree with what ought to be intuition. Just to be absolutely clear, the set that you've proposed is open. Right? $\endgroup$ – J. Min Dec 7 '17 at 0:06
  • $\begingroup$ Quite right. It is an open interval. We really need those to be open. $\endgroup$ – Matt Dec 7 '17 at 0:11
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If you accept that we have an intuitive understanding of what is meant by, for example, an open disc in $\Bbb R^2$, then your proposed definition will not work.

The boundary of such a disc is its bounding circle (apologies for tautology, but I'm talking about intuitive ideas, not formal calculations). So the disc and its boundary have empty intersection and the standard definition works. But the boundary is not empty, so your definition does not classify this "open disc" as an "open set".

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