Cardinality

[Nazia Tabassum]

What is the cardinality of the set A = {x:x is even and 2<X<17}? can help me anyone

 

[Deleted member 4993]

What is the cardinality of the set A = {x:x is even and 2<X<17}? can help me anyone

Please show us what you have tried and exactly where you are stuck.​

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[Dr.Peterson]

What is the cardinality of the set A = {x:x is even and 2<X<17}? can help me anyone

Cardinality means the number of elements in the set. So the first thing you might do is to just list out the elements of the set. (Then you can think about ways that would work better if 17 were replaced by, say, 1,000,017.)

 

[Nazia Tabassum]

Cardinality means the number of elements in the set. So the first thing you might do is to just list out the elements of the set. (Then you can think about ways that would work better if 17 were replaced by, say, 1,000,017.)

Can you please tell me the answer?

 

[tkhunny]

Can you please tell me the answer?

Yes, we are quite able to do that. What do YOU get?

 

[Dr.Peterson]

Just do what I suggested and show your thinking. Then we can discuss it.

 

[Nazia Tabassum]

If U= {1, 3, 7, 9, 11, 13}, then which are the subsets of U?

 

[Deleted member 4993]

If U= {1, 3, 7, 9, 11, 13}, then which are the subsets of U?

From your textbook (or classnotes or Google) please tell us:

the definition of a subset.

 

[Dr.Peterson]

If U= {1, 3, 7, 9, 11, 13}, then which are the subsets of U?

Have you abandoned the original question? This one is very different; the complete list will be fairly long, but counting them is easy.

 

[Steven G]

Although it is not your fault but U is a terrible name for a set.

 

[HallsofIvy]

Do you know what an "even number" is? Do you know what "cardinality of a set" means? I would hope so but there is no way to tell from your post. Can you list all the even numbers greater than 2 and less than 17? Then just count them!

There is a theorem that says that if a set has cardinality "s" then it has 2^s subsets. Here the set has 6 members, so cardinality 6, so there are 2^6= 64.

How do you write all 64 subsets without missing any? I recommend this:
First, list the empty subset which is a subset of every set: {}
Next, list all "singleton subsets": {1}, {3}, {7}, {9}, {11}, {13}.
Now list all two member sets with "1" as a member:
{1, 3}, {1, 7}, {1, 9}, {1, 11}, {1, 13}. There are 5 such sets.
Now list all two member sets with "3" as a member:
{3, 7}, {3, 9}, {3, 11}, {3, 13} (I did not include "{3, 2}" because "{2, 3}" was listed before. There are 4 such sets.
then all two member sets with "7" first (three sets), all two member sets with "9" first (two sets), and all two member sets with "11" first (one set).

So far there are: one set with 0 members, 6 sets with 1 member. and 5+ 4+ 3+ 2+ 1= 15 sets with 2 members.

Now, list all three member sets: {1, 3, 7}, {1, 3, 9}, {1, 3, 11}, {1, 3, 13}, {1, 7, 9}, {1, 7, 11}, {1, 7, 13}, {1, 11, 13}, {3, 7, 9}, {3, 7, 11}, {3, 7, 13} etc. (do you see the pattern?)