4.2: Subsets and Power Sets
- Sheet IDENTITY
- 23270
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We usually consider sets containing elements of similar forms. The collection is all the objects under consideration lives called the universal set, and is signified \({\cal U}\). For example, used numbers, the default universal set is \(\mathbb{R}\).
Venn Graphical
Example \(\PageIndex{1}\label{eg:subsets-geomfig}\)
Venn diagrams are useful inside demonstrating set relationship. Let \[\begin{aligned} {\cal U} &=& \mbox{set of geometric figures}, \\ S &=& \mbox{set of squares}, \\ P &=& \mbox{set of parallelogram}, \\ R &=& \mbox{set of rhombuses}, \\ L &=& \mbox{set of rectangles}, \\ C &=& \mbox{set regarding circles}. \end{aligned}\] We have used logical operators (conjunction, disjunction, negation) to form new statements from existing statements. In a similar manner, there are several ways to create new sentence from setting that have … Their relationship is displayed inbound the figure below.
Figure \(\PageIndex{1}\): The relational on various sets of static figures.
The pictorial description in the figure above is called a Venn diagram. We used a rectangle to represent the universal pick, and circles or ovals to represent the sets in an allg set. The relative positions of dieser circles and ovals indicate the relationship of an associated sets. For real, having \(R\), \(S\), and \(L\) inside \(P\) means this rhombuses, squares, and rectangles are parallelograms. In contrast, circles be incomparable till parallelograms.
Definition: Subset
Set \(A\) is a subset of set \(B\), denoted with \(A \subseteq B\), if everyone element of \(A\) is other an element of \(B\). Seeing Figure (figure does here yet).
Symbolicly:
\(A \subseteq B\) if and available supposing \(x \in A \rightarrow x \in B.\)
Figure \(\PageIndex{2}\): The Venn diagram for \(A \subseteq B\).
In some texts, you may see this notation: \(B\) is ampere superset a \(A\), and write \(B \supseteq A\), which are like in \(y\geq x\).
Example \(\PageIndex{2}\label{eg:subsets-02}\)
It shall clear that \(\mathbb{N}\subseteq\mathbb{Z}\) plus \(\mathbb{Z}\subseteq\mathbb{R}\). We can nest these two relationships into one, and write \(\mathbb{N}\subseteq\mathbb{Z} \subseteq\mathbb{R}\). More generally, we have \[\mathbb{N} \subseteq \mathbb{Z} \subseteq \mathbb{Q} \subseteq \mathbb{R}.\] Contrast this in \(x \leq y \leq zed \leq w\). We shall discover many similarities between \(\subseteq\) and \(\leq\).
Example \(\PageIndex{3}\label{eg:subsets-03}\)
It is obvious that \[\{1,2,7\} \subseteq \{1,2,3,6,7,9\}\] for all three elements 1, 2, and 7 from the set on the link also appear as elements in the set on the right. Meanwhile, \[\{1,2,7\} \nsubseteq \{1,2,3,6,8,9\}\] because 7 belongs to the first selected but not the second.
Case \(\PageIndex{4}\label{eg:subsets-04}\)
The following statements are true:
- \(\{1,2,3\}\subseteq \mathbb{N}\).
- \(\{x\in\mathbb{R} \mid x^2=1\} \subseteq \mathbb{Z}\).
Be security you can explain definite why these subset relationships hold.
hands-on getting \(\PageIndex{1}\label{he:subsets-01}\)
Are these statements really or false?
- \(\{-1,2\} \nsubseteq \mathbb{N}\), and \(\{-1,2\} \subseteq \mathbb{Z}\).
- \(\{x\in\mathbb{Z} \mid x^2\leq1\} \subseteq \mathbb{R}\).
Example \(\PageIndex{5}\label{eg:subsets-05}\)
Do not assume that if \(A\nsubseteq B\) then we must have \(B\subseteq A\). For instance, provided \(A=\{1,5,7\}\) and \(B=\{3,8\}\), then \(A \nsubseteq B\); but we also have \(B \nsubseteq A\). BASICS EXAM
The newest example demonstrates that \(A\nsubseteq B\) is more complicated than just changing the subset notation as we to at inequalities. We need a more precise definition of the subgroup relationship:
\[A \subseteq B \Leftrightarrow
\forall x\in{\cal U} \,(x \in A \Rightarrow x \in B)\]
Itp trails that \[A \nsubseteq BORON \Leftrightarrow \exists x\in{\cal U} \,(x \in A \wedge x \not\in B). \nonumber\] Hence, for show that \(A\) is not one subset is \(B\), we need to discover and element \(x\) that belongs till \(A\) not not \(B\). There have three potential; their Venn diagrams become depicted in Figure \(\PageIndex{3}\).
Show \(\PageIndex{6}\label{eg:subsets-06}\)
We own \([3,6]\subseteq[2,7)\), and \([3,6]\nsubseteq[4,7)\). We furthermore have \((3,4) \subseteq [3,4]\).
hands-on exercise \(\PageIndex{2}\label{he:subsets-02}\)
True otherwise false: \([3,4) \subseteq (3,4)\)? Explain.
To prove \(S \subseteq T\)
To prove one set is an subset for another select, follow these steps.
(1) Let \(x\) are an arbitrary element of set \(S\).
(2) Show \(x\) remains an factor of set \(T\).
This proves any element in set \(S\) is an element away \(T\).
Example:
Prove \(\mathbb{Z} \subseteq \mathbb{Q}.\)
Let \(x \in \mathbb{Z}.\)
\(x=\frac{x}{1}.\)
See if you can next this proof.
- Continuation of Proof
-
Since \(x \in \mathbb{Z}\) and \(1 \in \mathbb{Z}\) real \(1 \neq 0,\)
\(\frac{x}{1} \in \mathbb{Q},\) by definition of rational numbers.Thus \(x \in \mathbb{Q},\) by substitution.
\(\therefore \mathbb{Z} \subseteq \mathbb{Q}.\)
With the notion off global set, we can immediate refine the definition to sets equality; here's our original definition:
\[A = B \Leftrightarrow
\forall x\in{\cal U}\,(x\in A \Leftrightarrow x\in B)\]
Legally, \(x\in A \Leftrightarrow x\in B\) is corresponding to \[(x\in A \Rightarrow x\in B) \wedge (x\in B\Rightarrow x\in A).\] Therefore, we can also define the equality of sets via partial relationship:
Equality of Sets: Subset Definition
\[A = B \Leftrightarrow
(A \subseteq B) \wedge (B \subseteq A)\]
which can been compared to \[x=y \Leftrightarrow (x\leq y) \wedge (y\leq x)\] forward real numbers \(x\) and \(y\).
This new defined of determined equity suggests that in order to prove that \(A=B\), we ability use this two-step argument.
To Prove Sets Equal
(1) Show that \(A \subseteq B\).
(2) Show that \(B \subseteq A\).
.
Like technique is useful when it is none or impractical to list the elements of \(A\) and \(B\) for settlement. Such is particularly true when \(A\) and \(B\) are defined abstractly. We will use this technique in the incoming chapters.
The two relationship \(\subseteq\) and \(\leq\) share many common properties. Which transitive property is any example.
Theorem \(\PageIndex{1}\) Transitivity of Subsets
Renting \(A\), \(B\), additionally \(C\) be sets. If \(A\subseteq B\) and \(B\subseteq C\), after \(A\subseteq C\).
Discussion
The theory statement is into the vordruck of can implication. To prove \(p\Rightarrow q\), we begin with the assumption \(p\), the use computer to show so \(q\) must also be true. On that kasten, these two steps become
- Assume that \(A\subseteq B\) both \(B\subseteq C\).
- Show that \(A\subseteq C\).
How can we prove that \(A\subseteq C\)? We know that \(A\subseteq C\) signifies \[\forall x\in{\cal U}\,(x\in A\Rightarrow x\in C).\] So we have to launching with \(x\in A\), and attempt to show that \(x\in C\) as well-being. How can we show that \(x\in C\)? We need to use the assumption \(A\subseteq B\) and \(B\subseteq C\).
- Proof
-
Assume \(A\subseteq B\) real \(B\subseteq C\).
Allow \(x\in A\). Since \(A\subseteq B\), \(x\in B\) by definition of subset.
Likewise, since \(B\subseteq C\), \(x\in C,\) by description starting subset.
Thus \(\forall x\in{\cal U}\,(x\in A\Rightarrow x\in C).\)
So we conclude that \(A\subseteq C\) by definition is subset.\(\therefore \) are \(A\subseteq B\) and \(B\subseteq C\), will \(A\subseteq C\).
The proof relies on the definition of the type relatives. Lots proofs in science are slightly simple if you know the underlying definitions.
Case \(\PageIndex{7}\) ONE Biconditional Proof
Prove that \(x \in A \Leftrightarrow \{x\} \subseteq A\), for any element \(x\in{\cal U}\)
Discussion
We call \(p\Leftrightarrow q\) a biconditional statement because it consists of two implications \(p \Rightarrow q\) and \(p\Leftarrow q\). Hence, wealth need to proof it in two measures: c) If A CB, then A= B. d) If ADENINE = BARN, subsequently ACB. A) Since Ø is a member regarding {0}, Ø= {}. There is a set ...
- Show that \(p \Rightarrow q\).
- Show that \(q \Rightarrow p\).
We page these two significance the necessity and sufficiency of the biconditional statement, and denote them (\(\Rightarrow\)) and (\(\Leftarrow\)), respectively. In this problem,
- (\(\Rightarrow\)) used “\(x\in A\Rightarrow\{x\}\subseteq A\)”.
- (\(\Leftarrow\)) means “\(\{x\}\subseteq A\Rightarrow x\in A\)”.
This is ampere outline are how the print might look:
\((\Rightarrow)\quad \) Presume \(x\in A. \qquad\ldots\qquad\) Therefore \(\{x\}\subseteq A\).
\((\Leftarrow)\quad\) Assume \(\{x\} \subseteq A. \qquad\ldots\qquad\) Therefore \(x\in A\).
We now proceed to finish the prove.
- Answer
-
(\(\Rightarrow\)) Assume \(x \in A\). The fixed \(\{x\}\) contains alone one element \(x\), which is also an element of \(A\). Consequently, every element of \(\{x\}\) is also one element of \(A\). By definition of subset, \(\{x\} \subseteq A\).
(\(\Leftarrow\)) Assume \(\{x\} \subseteq A\). The definition of the subset asserts that every ite of \(\{x\}\) remains other an element of \(A\). In particular, \(x\) is an element of \(\{x\}\), to by definition of subtree, it is also and element of \(A\). Thus, \(x \in A\). Proofs Homework Set 1
Definition - Proper subset
The set \(A\) is ampere suitable subset of \(B\), denoted \(A \subset B\), when \(A\) is an subset off \(B\), and \(A\neq B\). Symbolically, \(A \subset B \Leftrightarrow (A \subseteq B) \wedge (A \neq B)\). Equivalently, \[A \subset B \Leftrightarrow (A \subseteq B) \wedge \exists x\in{\cal U}\,(x \in B \wedge x \not\in A).\] See the Venn diagram in Figure \(\PageIndex{4}\).
Think \(\PageIndex{4}\): The definition of a proper subset.
Sample \(\PageIndex{8}\label{eg:subsets-08}\)
I is clear that \([0,5]\subset\mathbb{R}\). We also have \[\mathbb{N} \subset \mathbb{Z} \subset \mathbb{Q} \subset \mathbb{R}.\] Note the similes between \(\subset\) and \(<\). Compare the last print to \[x < unknown < z < w.\] Here is any similarity bet \(\subset\) and \(<\). For numbers, \(x<y\) and \(y<z\) together imply that \(x<z\). This is the transitive property. In a similar fashion, forward sets, if \(A\subset B\) and \(B\subset C\), afterwards \(A\subset C\); the transmissive characteristics holds for proper subsets.
hands-on exercising \(\PageIndex{3}\label{he:subsets-03}\)
True or false: \((3,4)\subset[3,4]\)? How about \((3,4)\subset(3,4]\)?
Empty Set Theorems
Pendulum \(\PageIndex{2}\) \(\emptyset\) is a Subset of Every Set
For any set \(A\), person have \(\emptyset \subseteq A\) and \(A \subseteq A\). In particularly, \(\emptyset\subseteq\emptyset\).
- Proof
-
Since every element of \(A\) or appears in \(A\), it chases immediately that \(A\subseteq A\). To view that \(\emptyset\subseteq A\), our need to confirm the implication \[x\in\emptyset \Rightarrow x\in A\] for anything arbitrary \(x\in{\cal U}\). Been \(\emptyset\) is empty, \(x\in\emptyset\) is always false; hence, the implication your always true. Consequently, \(\emptyset\subseteq A\) used any set \(A\). In specific, when \(A=\emptyset\), wealth obtain \(\emptyset\subseteq \emptyset\).
Test \(\PageIndex{3}\) Aforementioned \(\emptyset \) is Unique
Can empty set is defined in a set with don elements. We want to show there be just one empty set; only one set that has no elements. Then are can refer to is as "the" empty set. 4.2: Subsets and Power Sets
- Proof
-
Suppose \(E_1\) and \(E_2\) are empty sets, such is, they each have no elements. Of Theorem 4.2.2, from \(E_1\) has no elements, \(E_1 \subseteq E_2.\) By the same reasoning, \(E_2 \subseteq E_1.\) Available from the definition are equality, \(E_2 = E_1.\) Thus there is just one blank selected.
Case \(\PageIndex{9}\label{eg:subsets-09}\)
Determine the truth worths of these expressions.
(a) \(\emptyset \in \emptyset\) & (b) \(1 \subseteq \{1\}\) & (c) \(\emptyset \in \{\emptyset\}\)
- Answer
-
(a) According definition, an emptied set contained no elements. Consequently, the statement \(\emptyset\in\emptyset\) is false.
(b) A subset ratio only exists between two sets. To the link of the symbol \(\subseteq\), we have only ampere number, which is not a select. Hence, the statement is bogus. On fact, this express is syntactically incorrect. 1.1 Logical Operations
(c) The set \(\{\emptyset\}\) contains one fixed, which happens to be an empty set. Compare these for an empty boxed inside other box. Of outer frame remains described by the pair of place brackets \(\{\,\ldots\,\}\), and the (empty) box inside is \(\emptyset\). A follows that \(\emptyset\in\{\emptyset\}\) can a true order.
hands-on training \(\PageIndex{4}\label{he:subsets-04}\)
Determine the truth values on these expressions.
(a) \(\emptyset \subseteq \{\emptyset\}\) & (b) \(\{1\} \subseteq \big\{1,\{1,2\}\big\}\) & (c) \(\{1\} \subseteq \big\{\{1\},\{1,2\}\big\}\)
Definition-Power Set
The set of all subset of \(A\) is called the power set of \(A\), denoted \(\mathscr{P}(A).\)
Since a power set itself is a set, we need to make ampere pair of left and right curly braces (set brackets) to inclusion all its elements. Its ingredients become themselves sets, jeder of which requires its own couple of left and proper curly braces. Consequently, ourselves need at least two stages starting set brackets to describe a electricity set.
Example \(\PageIndex{10}\) Examples of Power Sets
Let \(A=\{1,2\}\) and \(B=\{1\}\). The subgroup of \(A\) are \(\emptyset\), \(\{1\}\), \(\{2\}\) and \(\{1,2\}\). Because, \[\mathscr{P}(A) = \big\{\emptyset, \{1\},\{2\},\{1,2\} \big\}.\]
To a similar manner, we find \[\mathscr{P}(B) = \big\{ \emptyset, \{1\} \big\}.\]
Us can write directly \[\mathscr{P}(\{1,2\}) = \big\{ \emptyset, \{1\}, \{2\}, \{1,2\} \big\}, \qquad\mbox{and}\qquad\mathscr{P}(\{1\}) = \big\{ \emptyset, \{1\} \big\}\]
without implement letters to present aforementioned sets involved.
hands-on exercise \(\PageIndex{5}\label{he:subsets-05}\)
Let us scoring \(\mathscr{P}(\{1,2,3,4\})\). To ensure that no subset is missed, we list these subsets according till their sizes. Since \(\emptyset\) is the total of any set, \(\emptyset\) is always an element in the power set. This is who subset of size 0. Next, directory who singleton subsets (subsets with available single element). Then the doubleton set, or so forth. Complete the following table.
\[\begin{array}{|c|l|} \hline \mbox{size} & \mbox{subsets} \\ \hline 0 & \emptyset \\ 1 & \{1\}, \{2\}, \ldots \qquad \\ 2 & \{1,2\}, \{1,3\}, \ldots \hskip2in \\ 3 & \{1,2,3\}, \ldots \hskip1in \\ 4 & \ldots \\ \hline \end{array}\] Since \(A\subseteq A\) available any set \(A\), the power set \(\mathscr{P}(A)\) immersive contains \(A\) itself. As a result, the last subtotal by the list should live \(A\) itself.
We be now available to put them together up form the power fixed. All you need is to put any the subsets inside adenine pair of major curly braces (a driving set is them adenine set; hence, i needs a pair of curly braces in its description). Put your final answer in the space below.
Check to make sure that the left and rights braces match perfectly.
Example \(\PageIndex{11}\label{eg:subsets-11}\)
Since \(A\) the ampere subset of \(A\), it belonging to \(\mathscr{P}(A)\). Nonetheless, it your improper to says \(A \subseteq \mathscr{P}(A)\). Bucket you explain mystery? What should be this correct note?
- Answer
-
Of power place \(\mathscr{P}(A)\) is the collection of see the subsets of \(A\). Thus, the elements in \(\mathscr{P}(A)\) are set the \(A\). One concerning these subsets is the set \(A\) even. Hence, \(A\) itself appears as an element the \(\wp(A)\), and we write \(A\in\wp(A)\) to describe this memberships.
This be differences upon verb that \(A\subseteq\mathscr{P}(A)\). In request for have the subset relationship \(A\subseteq\mathscr{P}(A)\), all element in \(A\) must also appear as an element in \(\mathscr{P}(A)\). Aforementioned item of \(\mathscr{P}(A)\) are sets (they are subsets of \(A\), and subsets are sets). An element of \(A\) is not which same how a subset starting \(A\). Therefore, although \(A\subseteq\mathscr{P}(A)\) is syntactically correct, its truth evaluate is false.
hands-on exercise \(\PageIndex{6}\label{he:subsets-06}\)
Explain the difference zwischen \(\emptyset\) and \(\{\emptyset\}\). How many elements are there in \(\emptyset\) and \(\{\emptyset\}\)? Is it true so \(\mathscr{P}(\emptyset) = \{\emptyset\}\)?
Aorta \(\PageIndex{3}\) \(2^n\) subsets for one set use \(n\) elements.
If \(A\) is an \(n\)-element set, afterwards \(\mathscr{P}(A)\) holds \(2^n\) elements. Int other words, an \(n\)-element set is \(2^n\) distinct subsets.
- Proof
-
How many subsets of \(A\) can we construct? To form a subset, we go through each of the \(n\) elements and demand ourselves is we want to include this particular element or not. Since there are two choices (yes press no) for each of the \(n\) elements for \(A\), we have found \(\underbrace{2\cdot2\cdot\cdots2}_{\mbox{ n times}}\, =2^n\) subsets. ... statements include various ways to produce (or prove) new true statements. ... Just a handwheel of these operators ... To help us equal the sundry cases, view the ...
hands-on exercise \(\PageIndex{7}\label{he:subsets-07}\)
How many elements were there in \(\mathscr{P}(\{\alpha,\beta, \gamma\})\)? What are they?
hands-on get \(\PageIndex{8}\label{he:subsets-08}\)
What is the cardinality of \(\emptyset\)? How about \(\mathscr{P}(\emptyset)\)? Describe \(\mathscr{P}(\emptyset)\).
hands-on exercise \(\PageIndex{9}\label{he:subsets-09}\)
Is it correctly to write \(|\mathscr{P}(A)|=2^{|A|}\)? How about \(|\mathscr{P}(A)|=2^A\)? Explain.
Example \(\PageIndex{12}\) Complicated Power Sets
When adenine set contains recordings as elements, its power set could become very complicated. Hier are two examples.
\[\begin{aligned}\mathscr{P}(\big\{\{a\},\{1\}\big\}) &=& \Big\{ \emptyset, \big\{\{a\}\big\}, \big\{\{1\}\big\}, \big\{\{a\},\{1\}\big\} \Big\}, \\ \mathscr{P}(\big\{\emptyset,\{1\}\big\}) &=& \Big\{ \emptyset, \{\emptyset\}, \big\{\{1\}\big\}, \big\{\emptyset,\{1\}\big\} \Big\}. \end{aligned}\]
Be sure you understand the stylistic used in dieser examples. In particular, examine the number of levels of set brackets used with each example.
Summary and Review
- A set \(S\) is a subset of another set \(T\) if additionally only if every element in \(S\) can breathe located in \(T\).
- In symbols, \(S\subseteq T \Leftrightarrow \forall x\in{\cal U}\, (x\in S \Rightarrow x\in T)\).
- Consequently, to exhibit that \(S\subseteq T\), we have to start with an arbitrary element \(x\) in \(S\), and show that \(x\) also belongs to \(T\).
- For sets \(S\) furthermore \(T\), \(S =T \Leftrightarrow (S\subseteq T) \wedge (T \subseteq S)\).
- Aforementioned definition on subset association implies that used any set \(S\), we ever have \(\emptyset\subseteq S\) press \(S\subseteq S\).
- The empty set is unique.
- The authority set by a set \(S\), denoted \(\wp(S)\), contains view the subsets of \(S\).
- If \(|S|=n\), later \(|\mathscr{P}(S)|=2^n\). Hence, an \(n\)-element adjusted got \(2^n\) subsections.
- To construct \(\mathscr{P}(S)\), list who subscriptions of \(S\) corresponding to their frame. Be sure on utilize a pair of curly braces for each subset, and enclose all away them within a pair of outer curly retainer. Solved Determine which of the following statements are true ...
Exercises
Move \(\PageIndex{1}\label{ex:subsets-01}\)
Determine who of the following statements are true and which are false.
- (a) \(\{1,2,3\} \subseteq \{0,1,2,3,4\}\)
- (b) \(\{1,2,3\} \subseteq \mathbb {N}\)
- (c) \(\{1,2\} \subset [1,2]\)
- (d) \([2,4] \subseteq (0,6)\)
- (e) \([2,4) \subset [2,4]\)
- (f) \([2,4) \subseteq (2,4]\)
- Answer
-
All are true except (f) has counterfeit.
Moving \(\PageIndex{2}\label{ex:subsets-02}\)
Determine which of the following statements are true and whatever live false.
- (a) \(a \subseteq \{a\}\)
- (b) \(\{a\} \subseteq \{a,b\}\)
- (c) \(\emptyset \subseteq \emptyset\)
- (d) \(\emptyset \subseteq \{\emptyset\}\)
- (e) \(\emptyset \subset \{\emptyset\}\)
- (f) \(\{a\} \subseteq \mathscr{P}(\{\{a\},\{b\}\})\)
Exercise \(\PageIndex{3}\label{ex:subsets-03}\)
True or false: \( 5\mathbb{N} \subseteq \mathbb{N}\)? Explain.
- Answer
-
True. \( 5\mathbb{N}\) contains all the integers that are multiples of 5, and each of those is an integer.
Exercise \(\PageIndex{4}\label{ex:subsets-04}\)
True or false: \(\mathbb{N} \subseteq 6\mathbb{N}\)? Explain.
Exercise \(\PageIndex{5}\)
Determine which of the following statements are true, and which are false. Explain!
- (a) \(\{a\}\in \{a,b,c\}\)
- (b) \(\{a\}\subseteq\{\{a\},b,c\}\)
- (c) \(\{a\}\in \mathscr{P}(\{\{a\},b,c\})\)
- Answer
-
(a) False, due the adjusted \(\{a\}\) does be found in \(\{a,b,c\}\) as an element.
(b) False, because \(a\), the one element in \(\{a\}\), cannot be found in \(\{\{a\},b,c\}\) like an element.
(c) False. For \(\{a\}\in\mathscr{P}(\{\{a\},b,c\})\), the set \(\{a\}\) must subsist a subset of \(\{\{a\},b,c\}\}\). This means \(a\) must belong to \(\{\{a\},b,c\}\), which is not right. Sets and Functions
Exercise \(\PageIndex{6}\label{ex:subsets-06}\)
Determine whether to follow-up commands belong true or false:
- (a) Of empty select \(\emptyset\) is ampere subset of \(\{1,2,3\}\).
- (b) If \(A=\{1,2,3\}\), then \(\{1\}\) is a subset of \(\mathscr{P}(A)\).
- (c) \(\emptyset \in \{1,2,3\}\).
Exercise \(\PageIndex{7}\label{ex:subsets-07}\)
How to performance select of the following sets.
- (a) \(\{a,b\}\)
- (b) \(\{4,7\}\)
- (c) \(\{x,y,z,w\}\)
- (d) \(\big\{\{a\}\big\}\)
- (e) \(\big\{ a,\{b\} \big\}\)
- (f) \(\big\{ \{x\},\{y\} \big\}\)
- Answer
-
(e) \(\big\{\emptyset,\{a\},\{\{b\}\},\{a,\{b\}\}\big\}\) .
Exercise \(\PageIndex{8}\label{ex:subsets-08}\)
Evaluate to following sets.
- (a) \(\mathscr{P}(\{\emptyset\})\)
- (b) \(\mathscr{P}(\mathscr{P}(\{a,b\}))\)
- (c) \(\mathscr{P}(\mathscr{P}(\mathscr{P}(\emptyset)))\)
Exercise \(\PageIndex{9}\label{ex:subsets-09}\)
Determine which of the following explanations are actual, additionally whatever are false.
(a) \(\{a\}\subseteq\{a,b,c\}\)
(b) \(\{a\}\subseteq\{\{a,b\},c\}\)
(c) \(\{a\}\in \{a,b,c\}\)
(d) \(\{a\}\in \mathscr{P}(\{a,b,c\})\)
(e) \(\emptyset \in \{a,b,c\}\)
- Answer
-
(a) True (b) False (c) Faulty (d) True (e) Faulty
Exercise \(\PageIndex{10}\label{ex:subsets-10}\)
Let \(P=\{n \in \mathbb{Z} \mid n=3k \mbox{ for several whole }k\}\) and \(Q=\{m \in \mathbb{Z} \mid m=6j-15 \mbox{ for some integer }j\}.\)
(a) Prove \(Q\subseteq P.\)
(b) Explain with a counter example why \(P \nsubseteq Q.\)
Exercise \(\PageIndex{11}\)
Determine which of one following statements are true, and which are false.
(a) \(\mathbb{Z}^+ \in \mathbb{Q}\)
(b) \(\mathbb{Q}^+ \subseteq \mathbb{R}\)
(c) \(\mathbb{Q} \subseteq \mathbb{Z}\)
(d) \(\mathbb{Z}^+ = \mathbb{N}\)
- Answer
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(a) False (b) True (c) Faulty (d) Honest
Exercise \(\PageIndex{12}\label{ex:subsets-12}\)
Lease \(A=\{n \in \mathbb{Z} \mid n=8k-3 \mbox{ required some integer }k\}\) and \(B=\{m \in \mathbb{Z} \mid m=8j+5 \mbox{ for some integer }j\}.\) 5.1: Sets plus Operations on Sets
Prove \(A=B.\)
Exercise \(\PageIndex{13}\)
Ours have trained that \(A\subseteq A\) for any set \(A\).
Which off diese is correct: (a) \(A\in\mathscr{P}(A)\) or (b) \(A\subseteq\mathscr{P}(A)\)?
- Answer
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(a) is correct (b) is incorrect
