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For those of you who are already in the system (here's a link to the projects).
The "corrections" were not always correct. Somewhere on your page you should see a point total added to your score (perhaps as something like (18-10)/2, meaning that you got 10 originally, 18 on the revision, and you get half of those points).
Look over the key to see what went on....
One of the strange things was a set which contained the empty set as an element (but also has the empty set as a subset, as all sets do!): \[ C = \{ \emptyset,\{a,\{a\}\}\} \]
I think that some of you thought of that as "paradoxical" -- but it is not!
The empty set is an element of every power set.
Your power as a mathematician is to define things: if you don't like working with elements that look like \(\{a\}\) and \(\{b\}\), you can redefine them: \(A \equiv \{a\}\), and \(B \equiv \{b\}\); it kind of cleans things up.
unrestricted comprehension principle -- that for any sufficiently well-defined property, there is the set of all and only the objects that have that property.
Let's start with a well-defined property: define a regular set as one that is not an element of itself: \[ S \notin S \] and an irregular set as one which is an element of itself: \[ S \in S \] Consider the set of all regular sets (call it \(R\)). We will now show that \(R\) is neither regular nor irregular.
Suppose that \(R\) were regular; then \(R \notin R\). But that means that \(R \in R\), because \(R\) is regular.
Conversely suppose that \(R\) were irregular; then \(R \in R\). But that means that \(R \notin R\), as all sets are regular which are elements of \(R\).
Thus \(R\) is neither regular nor irregular. Or, as they put it in Wikipedia, \[ R = \{ x \mid x \not \in x \} \text{, then } R \in R \iff R \not \in R \]
and thus \(R\) fails the "unrestricted comprehension principle".
