Questions about subsets and members of the set {{b},{c},d,{∅}}

[Cabinbreaker]

For the set G where G = {{b},{c},d,{∅}}, I believe these are correct:

1) {{c}} \subseteq G
2) {b} \subseteq G
3) {d} \subseteq G
4) d \subseteq G
5) ∅ \subseteq G
6) c \notin \varphi(G)
7) {c} \in \varphi(G)
8) {b} \subseteq \varphi(G)
9) {{d}} \notin \varphi(G)
10) ∅ \in \varphi(G)
11) {∅} \in \varphi(G)
12) ∅ \subseteq \varphi(G)
13) {∅} \subseteq \varphi(G)
14) {{∅}} \subseteq \varphi(G)

If any are incorrect an explanation would be appreciated.

 

[haruspex]

For the set G where G = {{b},{c},d,{∅}}, I believe these are correct:

1) {{c}} \subseteq G
2) {b} \subseteq G
3) {d} \subseteq G
4) d \subseteq G
5) ∅ \subseteq G
6) c \notin \varphi(G)
7) {c} \in \varphi(G)
8) {b} \subseteq \varphi(G)
9) {{d}} \notin \varphi(G)
10) ∅ \in \varphi(G)
11) {∅} \in \varphi(G)
12) ∅ \subseteq \varphi(G)
13) {∅} \subseteq \varphi(G)
14) {{∅}} \subseteq \varphi(G)

If any are incorrect an explanation would be appreciated.

Two of the first 5 are wrong. If {x} is an element of G but x is not then {{x}} is a subset of G but {x} is not.

Is φ(G) supposed to be the power set? Not familiar with that notation. I am used to Ƥ (G) and 2^G. Assuming it is, X \subseteq G is the same as X \in \varphi(G), and it would also follow that {X} \subseteq \varphi(G). I think 4 of the last 9 are wrong.

 

[Cabinbreaker]

Yes, φ is the power set.

I have changed these to reflect what I believe is accurate but still have trouble with number 5.

G = {{b},{c},d,{∅}}

1) {{c}} ⊆ G
2) {b} ⊈ G
3) {d} ⊆ G
4) d ⊈ G
5) ∅ ⊆ G – would the empty set be a subset of G? I know {∅} is an element G, but ∅ is still a set.
6) c \in Ƥ(G)
7) {c} ∈ Ƥ(G)
8) {b} ⊈ Ƥ(G)
9) {{d}} ∉ Ƥ(G)
10) ∅ \notin Ƥ(G)
11) {∅} ∈ Ƥ(G)
12) ∅ ⊆ Ƥ(G)
13) {∅} ⊈ Ƥ(G)
14) {{∅}} ⊆ Ƥ(G)

 

[haruspex]

Yes, φ is the power set.

I have changed these to reflect what I believe is accurate but still have trouble with number 5.

G = {{b},{c},d,{∅}}

5) ∅ ⊆ G – would the empty set be a subset of G? I know {∅} is an element G, but ∅ is still a set.

1 to 4 are right now. The empty set is a subset of every set.

6) c ∈ Ƥ(G)

c is not even a set, as far as we know!

7) {c} ∈ Ƥ(G)

No, c is not an element of G, so {c} is not a subset of G, and so is not an element of the power set of G.

10) ∅ ∉ Ƥ(G)

Same qn as (5)

13) {∅} ⊈ Ƥ(G)

Reconsider in view of (5), (10)

14) {{∅}} ⊆ Ƥ(G)

No. This would have been true if the empty set had been an element of G.
∅ ⊆ X => ∅ ∈ Ƥ(X) => {∅} ⊆ Ƥ(X) (and these are true for any X)
Adding a level of {}:
∅ ∈ X <=> {∅} ⊆ X <=> {∅} ∈ Ƥ(X) <=> {{∅}} ⊆ Ƥ(X) (but ∅ ∉ G)
Adding another level of {}:
{∅} ∈ X <=> {{∅}} ⊆ X <=> {{∅}} ∈ Ƥ(X) <=> {{{∅}}} ⊆ Ƥ(X) (and these are true for G)

 

[SteveL27]

5) ∅ ⊆ G – would the empty set be a subset of G? I know {∅} is an element G, but ∅ is still a set.

As haruspex mentioned, ∅ is a subset of every set. To see why this is this, recall the definition of "subset." If A and B are sets, we say that A is a subset of B if

x \in A implies x \in B

Now this is an example of one of those "empty set arguments" that take some getting used to, but that you'll see over and over as you learn set theory.

I claim that if G is any set, x \in ∅ implies x \in G. Since there is no x \in ∅, it's vacuously true that x \in ∅ implies x \in G.

In other words, you can't show me an x that's an element of the empty set that's not also an element of G. That's because you can't show me any element of the empty set at all!

Therefore, formally, we do have the truth of the statement "x \in ∅ implies x \in G"

So by definition, ∅ is a subset of G, where G is any set at all.

It's worth working through this argument till it makes sense to you; because vacuous arguments involving the empty set come up all the time.

Also it's a good example of how referring directly back to the actual definition is often the best way to approach these problems.

 

[Cabinbreaker]

Thank you for the help! I found this most helpful in clearing up most of my misunderstandings!

No. This would have been true if the empty set had been an element of G.
∅ ⊆ X => ∅ ∈ Ƥ(X) => {∅} ⊆ Ƥ(X) (and these are true for any X)
Adding a level of {}:
∅ ∈ X <=> {∅} ⊆ X <=> {∅} ∈ Ƥ(X) <=> {{∅}} ⊆ Ƥ(X) (but ∅ ∉ G)
Adding another level of {}:
{∅} ∈ X <=> {{∅}} ⊆ X <=> {{∅}} ∈ Ƥ(X) <=> {{{∅}}} ⊆ Ƥ(X) (and these are true for G)