-2

If I have a set A with |A| = n then what would be the cardinality of the following set?

{ | ∈ (A), |X| ≤ 1}

What I understand is that this is the set of all X such that it is part of the power set of A and cardinality is less than or equal to 1. Thus I say the cardinality of this set is 2. Because there are two sets: {}, the empty set and {X} which are part of the power set of A.

Am I understanding correctly?

E. Ron
  • 3
  • 2
  • No. You are not understanding correctly. $X \ne {X}$ and ${X}$ is (probably) not a subset of $A$ and so is not an element of $P(A)$. And if you think about it if $a,b,c \in A$ then ${a}\subset X$ and $|{a}| = 1$ and ${b}\subset X$ and $|{\a}|=1$. So ${a},{b},\emptyset \in {X|X\in P(A),|X|\le 1}$ so the cardinality is at least $3$ if $n\ge 2$. – fleablood Sep 23 '20 at 20:17

2 Answers2

0

You have $1$ subset of $A$ of cardinality $0$ (the empty subset).

And you have $n$ subsets of $A$ of cardinality $1$ (each $\lbrace x \rbrace$ for $x \in A$).

So the set of subsets of cardinality at most $1$ has cardinality $$n+1 $$

TheSilverDoe
  • 29,720
  • so the X represents all elements of A? I am just confused by how { | ∈ (A) = n. I know that A has n different elements, and for each there is one subset in the powerset, thus n subsets. But what does X represent then? Thanks. – E. Ron Sep 23 '20 at 20:16
  • $X$ is a variable. It just means an unspecified subset of $A$. If $A={a,b,c}$ so $|A| =3$ then there are $8$ subsets of $A$. $X$ can be any one of those subsets. ${X|X\in P(A) ;|X|\le 1} = {X| X = \emptyset, {a},{b},{c}{a,b},{a,c},{b,c},$ or ${a,b,c}$ so that $|X|=0$ or $1}={X|X = \emptyset, {a},{b}, $ or ${c}}= {\emptyset, {a},{b},{c}}$. – fleablood Sep 23 '20 at 20:24
  • @fleablood Thanks, makes sense now. – E. Ron Sep 23 '20 at 20:28
0

You seem to be have trouble with set notation. $X$ is a place holder variable that is not any actual set. It merely means "Let $X$ be any element of $P(A)$" which is to say "Let $X$ be any subset of $A$".

So $\{X| X\in P(A); |X| \le 1\} =$ the set of all subsets of $A$ that have a cardinality of at most one.

And the question "what is the cardinality of $\{X| X\in P(A); |X| \le 1\}$" is asking nothing more or less than "how many subsets of $A$ have one or fewer elements".

And the answer to that is obvious. The empty set is a subset with zero elements, and every set with a single element has one element. And any set with more than one element has a cardinality of more than one. As there are $n$ elements in $A$ there are $n$ sets with a single element. Adding the empty set we have the cardinality is $n+1$.

Another way of looking at this is:

If $A = \{a_1,a_2,......, a_n\}$ then $\{X| X\in P(A); |X| \le 1\}=\{\emptyset, \{a_1\},\{a_2\},....., \{a_n\}\}$

fleablood
  • 124,253