Thank you for these lectures; I found them very useful! Also I really appreciated the way you suggest to remember wich is "finite" and wich "arbitrary" between intersection and union in the definition of open and closed sets.
@8:25, the intersection of Z and Z complement is written not empty, but it should be empty. However, the instructor does say it's empty so it really doesn't matter.
ya intersection of (Z and Z complement) are empty . but as it is given the point z is arbitrarily close to Z, so from the above definition then the open set (Z complement) containing z intersection the given set (Z) is not empty... here it contradicts.
@@somi11 Thank you, now this makes sense! I was also confused because he said he proves it by a "Contrapositive" and I assumed he meant the Contraposition. So to be clear, the proof is essentially: Let Z^c be open and z an arbitrary close point to Z. Assume z is not in Z. Since z not in Z, it is in Z^c. And since Z^c is open, it is an open neighbourhood of z. By definition of arbitrary close points, the intersection of Z with Z^c is therefor not empty. This is a contradiction, since the intersection of any set with its complement is empty. Therefor z must be in Z.
In the proof of Prop-Defn 1, the intersection of Z and its complement should of course be empty. On the board, it is written that the intersection is not equal to the empty set.
@@DanielChanMaths I had this "something's wrong here" feeling during this part of the video lol, but I eventually saw it after a few passes. Great content btw :)
Thank you for these lectures; I found them very useful! Also I really appreciated the way you suggest to remember wich is "finite" and wich "arbitrary" between intersection and union in the definition of open and closed sets.
@8:25, the intersection of Z and Z complement is written not empty, but it should be empty. However, the instructor does say it's empty so it really doesn't matter.
ya intersection of (Z and Z complement) are empty
. but as it is given the point z is arbitrarily close to Z, so from the above definition then the open set (Z complement) containing z intersection the given set (Z) is not empty... here it contradicts.
@@somi11 Thank you, now this makes sense! I was also confused because he said he proves it by a "Contrapositive" and I assumed he meant the Contraposition. So to be clear, the proof is essentially:
Let Z^c be open and z an arbitrary close point to Z. Assume z is not in Z.
Since z not in Z, it is in Z^c. And since Z^c is open, it is an open neighbourhood of z.
By definition of arbitrary close points, the intersection of Z with Z^c is therefor not empty.
This is a contradiction, since the intersection of any set with its complement is empty.
Therefor z must be in Z.
Thank you for explaining this so clearly. I wasn't able to find the mistake on the board though
In the proof of Prop-Defn 1, the intersection of Z and its complement should of course be empty. On the board, it is written that the intersection is not equal to the empty set.
@@DanielChanMaths I had this "something's wrong here" feeling during this part of the video lol, but I eventually saw it after a few passes.
Great content btw :)
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