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I'm not sure how to answer this question:

Construct a set $A$, which is a subset of $[0, 1] \times [0, 1]$, such that $A$ contains at most one point on the horizontal and vertical lines, but boundary $(A) = [0, 1]\times [0, 1]$.

Is the answer just the open rectangle $(0, 1) \times (0, 1)$? How do you prove that the boundary of an open rectangle is a closed rectangle?

Thanks.

Michael Albanese
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william
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1 Answers1

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The boundary of the open rectangle $(0,1)\times(0,1)$ is not $[0,1]\times[0,1]$. It is $\{0\}\times[0,1]\cup\{1\}\times[0,1]\cup[0,1]\times\{0\}\cup[0,1]\times\{1\}$. Also the open rectangle contains far more than one point on each horizontal and vertical line excluding the boundary lines.

To construct the desired set $A$ first pick a point with both coordinates being rational and in $[0,1]\times[0,1]]$. Subdivide $[0,1]\times[0,1]$ into four squares by forming the lines $\{\frac{1}{2}\}\times[0,1]$ and $[0,1]\times\{\frac{1}{2}\}$. Pick from each square an element such that both coordinates are rational and such that they don't lie a horizontal line or vertical line that can connect them to the previously chosen points. Continue subdividing into quarters as before and repeating the procedure. This process can be continued infinitely many times since there are infinitely many rationals and since at any finite stage we have only made finitely many choices and hence infinitely many lines are still available. The resulting set is dense by construction and has no interior since it is totally disconnected. The boundary will have the desired property.

user71352
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  • Damn, I was hoping for an answer I understood, but I really appreciate it anyway. Your method looks like the 'hint' on my question sheet. I only just began analysis and despise it already. – william Jul 23 '14 at 06:00
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    To understand a complicated description like this it is usually best to draw a picture. I find analysis is best understood through pictures. I would have included one put I couldn't find a picture online that described it perfectly. If there is any way in which I can help you further understand I would be happy to. – user71352 Jul 23 '14 at 06:04
  • Wow thank you dearly. Actually this course is called 'multivariable calculus' so I can't understand why we're doing this stuff. With your subdividing method, what do you end up with eventually? A set containing infinite points that's all inside of [0, 1] x [0,1]? – william Jul 23 '14 at 06:31
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    Yes it will be just an infinite collection of points in $[0,1]\times[0,1]$. An explicit description of $A$, as constructed above, may be difficult to get since I am implicitly enumerating the rationals. However, for your purposes the exact set is not important. What is important is the properties which it has. Notice that any neighbourhood of any point in $[0,1]\times[0,1]$ intersects $A$ (by density) in a countable set but balls contain uncountably many points. Hence, any neighbourhood of any point in $[0,1]\times[0,1]$ belongs to the boundary of $A$. – user71352 Jul 23 '14 at 06:48