00:01
Hi, here we have an exercise in logic.
00:04
We have that f -a be the predicate that a is a finite set.
00:10
S -a -coma b the predicate that a is containing b.
00:16
Suppose the universe of discourse consists of all sets, translate the statement into cibald.
00:22
So we have four statements here and we start with statements with a first statement.
00:28
Let me just zoom in.
00:30
Here so you can see better so the first one says that not all sets are finite and that means so if i take for example you to be there well it cannot be the set of all sets because there is not such a thing but let's see we have a power set so what we are going to have is that there is a set set in u that it is not finite that means that does not satisfy f of a so there is a in u not f of a then the second it says every subset of a finite set is finite so what we do is we have that for if i take any set that it is finite so that i have f of a and then i have also end.
01:39
I have that a set b is a subset of a.
01:45
So if i have these two together, that means a finite set and any subset of a finite set, then these two together imply that the set b is finite.
02:01
Here i don't have any restrictions.
02:04
I took any set that it is finite, i took any subset of that set, and then these two together, a is finite, b is a subset of a implies that b is finite.
02:19
Okay, so this is part two, part three.
02:25
In order to keep my screen zoomed, i'm going to, i will need to erase the first answer.
02:32
So let's see what is part 3 what we have here for part 3 it says no infinite set is contained in a finite set so what i do i take a set a i take a set a that it is not finite so what i have here not so this says that a is not finite and then we have that not infinite set is containing a finite set and we have another set b that it is finite...