Shave and a Haircut!

Gabriellebirchak/ April 26, 2020/ Contemporary History, Late Modern History, Modern History, Uncategorized

First there was …

Then came along …

The fol­low­ing expla­na­tion comes from one of my absolute favorite col­lege text­books (and per­son­al math bibles) Dis­crete Math­e­mat­ics with Appli­ca­tions by Dr. Susan­na S. Epp.

For the most part, sets are not ele­ments of them­selves. A per­fect exam­ple would be to say that the set of all cats is not a cat, which helps to explain that the set of all inte­gers is not an inte­ger. How­ev­er, in some cas­es, sets are ele­ments of them­selves. So, to explain, the set of all salt par­ti­cles is salt par­ti­cles, which helps to explain that the set of all abstract ideas is an abstract idea.

So, to explain Russell’s paradox

S=\lbrace N |\space N \space  \text {is a set and} \space N \notin N \rbrace

The ques­tion then, is S an ele­ment of itself? In oth­er words, Does the bar­ber shave himself?

The answer is Nei­ther yes nor no. If the bar­ber shaves him­self, then he’s a mem­ber of the group of men who shave themselves. 

 \text {If} \space S\in S, \text {then} \space S \space \text {satifies the defining property for}  \space S,  \text{therefore}  \space S\notin S. 

 \text {But if}  \space S\notin S, \text {then}  \space S  \space \text {is a set such that} \space S\notin S,  

 \text {and}  \space S  \space \text {satisfies the defining property for}  \space S. 

 \text {This implies that}   \space S\in S.  

\text{Therefore, neither is}  \space S\in S, \text {nor is} \space S\notin S. 

This is a contradiction. 

Russell’s para­dox, how­ev­er, is no longer con­tra­dic­to­ry when we use a pred­i­cate as a defin­ing prop­er­ty and spec­i­fy that the set is a sub­set of a known set.

By doing so, we remove the prop­er­ty “the set of all sets that are not ele­ments of them­selves.” As a result, we only address the prop­er­ty that “the set of all sets that are sub­sets of some known set and that are not ele­ments of themselves.” 

Now it’s not a contradiction.

\text {Let} \space L \space \text {be a set of sets and let} \space S=\lbrace N \in L|N \notin N \rbrace. \space \text {Is} \space S\in S?

No, it is not. 

\text {If} \space S \in S, \text {then} \space S \space \text {satisfies the defining property for} \space S.  

\text {Therefore,} \space S \notin S.

\text{When the definition of} \space S \space \text {is changed, only}  

S \in S \space --> \space S \notin S \space \text{can be proved.} 

\text {Therefore} \space S\notin L.

…which is not contradictory. 

But THEN along came.….

…who showed us all that it is not pos­si­ble to prove in a math­e­mat­i­cal­ly rig­or­ous way that math­e­mat­ics is free of contradictions!