Set Theory (Pure Basics)

by kamatadu, May 21, 2022, 6:16 PM

Definitions that I felt might be important

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DEFINITION 1:

Definition 1 wrote:

Intervals: $(a, b)$ is an Open-Interval, $[a, b]$ is a Closed-Interval, $[a, b)$ and $(a, b]$ are both Half-Closed-Intervals/Half-Open-Intervals.
$\begin{align*} x \in (a, b) \implies a < x < b \text{ and } x \in \mathbb{R}\\ x \in [a, b] \implies a \le x \le b \text{ and } x \in \mathbb{R}\\ x \in [a, b) \implies a \le x < b \text{ and } x \in \mathbb{R}\\ x \in (a, b] \implies a < x \le b \text{ and } x \in \mathbb{R}. \end{align*}$
Example: $x \in (2, 3) \implies 2 < x < 3$ and $x \in \mathbb{R}$ .

Null Set: A null set is a set $^{*}$ which has no elements in it.

Example: $\phi = \{ x \mid x \in (2, 3) \land x \in \mathbb{N} \} = \{\}$ .

Subset: $A \subseteq B$ implies that each element of $A$ is contained in $B$ , i.e., if $x \in A$ , then $x \in B$ .

Example: For, $A = \{2, 3, 5\}$ and $B = \{1, 2, 3, 4, 5\}$ , $A \subseteq B$ .

Equal Sets: $A = B$ , if and only if the both the contents of the two sets are exactly the same, i.e, $A = B$ if and only if every element of $A$ is present in $B$ and every element of $B$ is present in $A$ . In other words, $A = B$ if and only if $A \subseteq B$ and $B \subseteq A$ .

Example: For, $A = \{ 1, 2, 3 \}$ and $B = \{ 2, 1, 3 \}$ , $A = B$ .

Proper Subset: $A \subset B$ , implies that $A \subseteq B$ , but $A \not= B$ , i.e, $A$ is a subset of $B$ but there exists at least one element in $B$ that does not exist in $A$ .

Example: For, $A = \{2, 3, 4, 5\}$ and $B = \{1, 2, 3, 4, 5\}$ , $A \subset B$ because $A \subseteq B$ and $1 \in B$ , but $1 \not\in A$ .

Universal Set: $U$ is the set of which all the sets in the current context are a subset of.

Example: For, $A = \{ 1, 2, 3, 4 \}$ , $U$ can be the set of integer.

Power Set: The Power Set of a given set $A$ is the set of all its subsets. It is denoted by $\mathcal{P}(A)$ . In other words, $\mathcal{P}(A) = \{X \mid X \subseteq A\}$ .

Example: For, $A = \{1, 2, 3\}$ , $\mathcal{P}(A) = \{\phi, \{1\}, \{2\}, \{3\}, \{1, 2\}, \{2, 3\}, \{3, 1\}, \{1, 2, 3\}\}$ .

Union of Sets: $A \cup B$ is the set of all elements which either belong to $A$ or to $B$ or to both $A$ and $B$ . In other words $A \cup B = \{x \mid x \in A \lor x \in B\}$ .

Example: For, $A = \{ 2 , 3, 5 \}$ and $B = \{ 3, 5, 6, 7 \}$ , $A \cup B = \{ 2, 3, 5, 6, 7 \}$ .

Intersection of Sets: $A \cap B$ is the set of all elements which are common to $A$ and $B$ , i.e., it is the set of all elements that belong to both $A$ and $B$ . In other words $A \cap B = \{x \mid x \in A \land x \in B\}$ .

Example: For, $A = \{ 2 , 3, 5 \}$ and $B = \{ 3, 5, 6, 7 \}$ , $A \cap B = \{ 3, 5 \}$ .

Disjoint Sets: $A$ and $B$ are disjoint if they have no elements in common. In other words, $A$ and $B$ are disjoint if $A \cap B = \phi$ .

Example: For, $A = \{ 1, 2, 3, 4 \}$ and $B = \{ 6, 7, 8 \}$ , $A$ and $B$ are disjoint.

Difference of Sets: $A-B$ is the set of all elements which belong to $A$ but do not belong to $B$ . In other words, $A - B = \{ x \mid x\in A \land x \not\in B \}$ .

Example: For, $A = \{2, 3, 4, 5, 6\}$ and $B = \{ 5, 6, 7, 8 \}$ , $A - B = \{ 2, 3, 4 \}$ .

Complement of a Set: $A^C$ w.r.t its Universal Set $U$ is the set of all elements which belong to $U$ but do not belong to $A$ . In other words $A^C = U - A = \{ x \mid x \in U \land x \not\in A \}$ .

Example: For $U = \{ 1, 2, 3, 4,5,6,7,8,9 \}$ and $A = \{ 1, 2, 3, 5, 7 \}$ , $A^C = \{4,6,8,9\}$ .

$^{*}$ Again, $\phi$ represents an empty set, not just an element :read:

Now here are a few basic lemmas that help us make the definitions more firm

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LEMMA 1:

Lemma 1 wrote:

Lemma 1.1: $\phi \subseteq A$ for any set $A$ .

Proof 1.1

Lemma 1.2: $A \subseteq A$ for any set $A$ .

Proof 1.2

Lemma 1.3: If $A \subseteq B$ and $B \subseteq C$ , then $A \subseteq C$ .

Proof 1.3

Lemma 1.4: $A \subseteq B$ and $B \subseteq A$ if and only if $A = B$ .

Proof 1.4

Lemma 1.5: If $A$ is a finite set and contains $n$ elements, then $\mathcal{P}(A)$ has $2^n$ elements.

Proof 1.5

Now for the part of Theorems of Algebra of Sets :wallbash:

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LAWS OF ALGEBRA OF SETS:

Laws of Algebra of Sets wrote:

Idempotent Laws: For any set $A$ , we have:

$A \cup A = A$
$A \cap A = A$ .

Commutative Laws: For any two sets $A$ and $B$ , we have:

$A \cup B = B \cup A$
$A \cap B = B \cap A$ .

Associative Laws: For any three sets $A$ , $B$ and $C$ , we have:

$A \cup (B \cup C) = (A \cup B) \cup C$
$A \cap (B \cap C) = (A \cap B) \cap C$ .

Distributive Laws: For any three sets $A$ , $B$ and $C$ , we have:

$A \cup (B \cap C) = (A \cup B) \cap (A \cup B)$
$A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$ .

De Morgan's Laws: For any two sets $A$ and $B$ , we have:

$(A \cup B)^C = A^C \cap B^C$
$(A \cap B)^C = A^C \cup B^C$ .

Identity Laws: Let $U$ be the universal set, $\phi$ the null set and $A$ be any subset of $U$ , then we have:

$A \cup U = U$
$A \cap U = A$
$A \cup \phi = A$
$A \cap \phi = \phi$ .

Complement Laws: Let $U$ be the universal set, $\phi$ the null set and $A$ be any subset of $U$ , then we have:

$A \cup A^C = U$
$A \cap A^C = \phi$
$(A^C)^C = A$
$U^C = \phi$
$\phi^C = U$ .

REMARK 1:

Remark 1 wrote:

For any set $A$ and $B$ with their Universal Set $U$ and $\phi$ the null set, we have:

$A \cup A = A$
$A \cap A = A$
$A \cup U = U$
$A \cap U = A$
$A \cup \phi = A$
$A \cap \phi = \phi$
$A \subseteq A \cup B$
$A \cap B \subseteq A$
$A \cup A^C = U$
$A \cap A^C = \phi$
$(A^C)^C = A$
$A = U - A^C$
$U^C = \phi$
$\phi ^ C = U$
$A - B^C = A \cap B$
$A \cap (B - C) = (A \cap B) - (A \cap C)$
$A - (B \cup C) = (A - B) \cap (A - C)$
$A - (B \cap C) = (A - B)\cup (A - C)$
$(A \cup B) - C = (A - C) \cup (B - C)$
$(A \cap B) - C = (A - C) \cap (B - C)$ .

REMARK 2:

Remark 2 wrote:

For any set $A$ and $B$ , we have:
$\begin{align*} x \in (A \cup B) \implies x \in A \lor x \in B.\\ x \in (A \cap B) \implies x \in A \land x \in B.\\ x \not\in (A \cup B) \implies x \not\in A \land x \not\in B.\\ x \not\in (A \cap B) \implies x \not\in A \lor x \not\in B. \end{align*}$

Note: The last $6$ points of REMARK 1 and all of REMARK 2 are especially powerful for proving the basic Theorems and Lemmas of Set Theory using Set Algebra :stretcher:

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LEMMA 2:

Lemma 2 wrote:

Lemma 2.1: If $A$ and $B$ are two finite sets, then:
$|A \cup B| = |A| + |B| - |A \cap B|.$
Proof 2.1

Lemma 2.2: If $A$ , $B$ and $C$ are three finite sets, then:
$|A \,\cup \,B \,\cup \,C| = |A|\, + \,|B| \,+ \,|C|\, - \,|A \,\cap\, B| \,-\, |B \,\cap \,C|\, -\, |C \,\cap\, A|\, + \,|A \, \cap \,B\, \cap \,C|.$
Proof 2.2

Now moving onto the "final" section, VENN DIAGRAMS 1

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VENN DIAGRAMS 1:

Venn Diagrams 1 wrote:

For this section, I personally think that it is enough to look at the proof of Lemma 2.1 from the previous section

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Lemma 2.1: If $A$ and $B$ are two finite sets, then:
$|A \cup B| = |A| + |B| - |A \cap B|.$
Proof: Proof without words

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Here are some problems to wrap up this post

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PROBLEMS 1:

Problems 1 wrote:

Problem 1.1: Determine whether the following are true or not:

For $A = \{1, 2, 3\}$ and $B = \{ 1, 2, 3, 4, 5 \}$ , is $A \subseteq B?$
For $A = \{1, 2, 3, 4\}$ and $B = \{ 1, 2, 3, 5 \}$ , is $A \subseteq B?$
For $A = \{1, \{2, 3\}\}$ and $B = \{ 1, 2, 3, 4, 5 \}$ , is $A \subseteq B?$
For $A = \{\{2, 3\}\}$ and $B = \{ 1, 2, 3, 4, 5 \}$ , is $A \subseteq B?$
For $A = \{1, \{2, 3\}\}$ and $B = \{ 1, \{2, 3\}, 4, 5 \}$ , is $A \subseteq B?$

Soln 1 Soln 2 Soln 3 Soln 4 Soln 5

Problem 1.2: For any three sets $A$ , $B$ and $C$ , using Set Algebra, prove that:

$A \cap (B - C) = (A \cap B) - (A \cap C)$
$A - (B \cup C) = (A - B) \cap (A - C)$
$A - (B \cap C) = (A - B) \cup (A - C)$ .

Hint 1

Problem 1.3: For any two sets $A$ and $B$ , prove that $A = B$ if and only if $\mathcal{P}(A) = \mathcal{P}(B)$ .

Hint 1 Hint 2 Hint 3

Problem 1.4: For any two sets $A$ and $B$ , prove that:

$\mathcal{P}(A) \cup \mathcal{P}(B) \subseteq \mathcal{P}(A \cup B)$
$\mathcal{P}(A \cap B) = \mathcal{P}(A) \cap \mathcal{P}(B)$ .

Hint 1

Problem 1.5: For any two sets $A$ and $B$ , prove that: $(A \cup B) - (A \cap B) = (B - A) \cup (A - B).$

Problem 1.6: For any two sets $A$ and $B$ , using Set Algebra, prove that:

$A - B = B^C - A^C$
$(A \cup B) \cap (A \cup B^C) = A$
$(A \cup B) \cap A = A$
$A \cap (B - A) = \phi$ .

Problem 1.7: For any three sets $A$ , $B$ and $C$ , using Set Algebra, prove that:

$(A \cup B) - C = (A - C) \cup (B - C)$
$(A \cap B) - C = (A - C) \cap (B - C)$ .

Problem 1.8: In an examination, $45\%$ of the candidates have passed in English, $40\%$ have passed in Literature, while $30\%$ have passed in both the subjects. Find the total number of candidates if $90$ of them have failed in both the subjects.

Hint 1

Problem 1.9: In a survey concerning the smoking habits of consumers it was found that $50\%$ smoke cigarette $A$ , $45\%$ smoke $B$ , $40\%$ smoke $C$ , $25\%$ smoke $A$ and $B$ , $10\%$ smoke $B$ and $C$ , $16\%$ smoke $C$ and $A$ , $8\%$ smoke all the three brands. What percentage:

do not smoke
smoke only $A$ brand
smoke exactly two brands of cigarettes?

Hint 1

Problem 1.10: A factory inspector examined the defects in hardness, finishing and dimensions of an item. After examining $100$ items he gave the following report:

All three defects - $5$ , defect in hardness and finishing - $10$ , defect in dimensions and finishing $8$ , defect in dimensions and hardness - $20$ . Defect in finishing $30$ , in hardness - $23$ and in dimensions - $50$ . The inspector was fined why?

Hint 1 Hint 2

WOOF!!

. That's all for this post :rotfl:

. See you again in the new post again! :bye:

Tysm for reading this post

. Oh and btw, for the people who know me from a few months back, if u even for once thought what is this dude doing, why is he doing all these things instead of MO. Unfortunately in not that much into MO as before as I stopped doing geo (the only thing I can :")) SAR plix halp this nub :"), u ORSMAX plix gib teeps :"), plix halp this NUMBAX.

:omighty:

ORSMAX SAR.

This post has been edited 49 times. Last edited by kamatadu, Jun 27, 2022, 6:45 PM
Reason: asdf

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3 Comments

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Comment if you thought $\mathcal{P}(A)$ was polar of $A$ :yaway:

by BVKRB-, May 24, 2022, 4:45 AM

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Sar whai u stop posting sar

by BVKRB-, Jun 7, 2022, 8:53 AM

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@above sori sar, me continuing to post again sar

by kamatadu, Jun 7, 2022, 6:08 PM

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orz saar
by Levieee, Apr 16, 2025, 9:02 PM
i feel like i see a shoutout
by Nuterrow, Feb 27, 2025, 4:51 PM
Thanks for the tips
by math_holmes15, Feb 26, 2025, 6:13 AM
Good Luck with your new series
by SomeonecoolLovesMaths, Feb 24, 2025, 11:38 AM
hawsoworzz plij gibb teeps on how to attains skeels
by L13832, Nov 25, 2024, 6:53 AM
E bhai hothat cs change korli ki byapar
by HoRI_DA_GRe8, Nov 3, 2024, 8:03 PM
Ngl that pfp is orz
by factcheck315, Nov 2, 2024, 9:09 PM
finally shouting out orz ppl in orz blog
by Kappa_Beta_725, Oct 25, 2024, 6:13 PM
Omega lavil
Pro lavil
God lavil
Lavil lavil

by Sreemani76, Apr 19, 2024, 7:20 PM
Nice
by Om245, Jan 12, 2024, 2:18 AM
@below We lost finals, so GG...
by kamatadu, Nov 26, 2023, 2:43 PM
We won semis
by Project_Donkey_into_M4, Nov 15, 2023, 6:05 PM
yess sss
by polynomialian, Nov 3, 2023, 3:10 PM
We had a great camp in ISI back in February for MTRP (MATHEMATICS TALENT REWARD PROGRAM). So since the fest season of CMI and isi is near,this means the season is back again!!!.So should we post the mtrp 2023 experience?

Comment!
by HoRI_DA_GRe8, Nov 2, 2023, 4:27 PM
Saar good luck for RMO
haw prep going btw
Me dead non geo :skull:
by BVKRB-, Oct 20, 2023, 2:06 PM
ooo you also trying sharygin poblams nais saar, tell progress lol
by BVKRB-, Jul 5, 2022, 5:36 AM
@orsmax_viewer_sar, me nubmax is trying the sharygin poblams so me numbax not get time to update blog for few days :")
by kamatadu, Jul 3, 2022, 5:20 PM
@below no sar, I waste too much time sar :")
by kamatadu, Jul 1, 2022, 3:25 PM
hello sar bhul gaye kya
by anurag27826, Jun 28, 2022, 2:19 PM
AT LAST!!! Post on SET THEORY is finally complete
by kamatadu, Jun 21, 2022, 3:15 PM
IM BACK! Check the latest post for the description of why I went missing
by kamatadu, Jun 15, 2022, 6:09 PM
Guys im really sorry for not updating the blog for so long, actually ive been busy with fixing my mobile since the past few days, so I couldnt update my blog, I believe I will be done with this work by tomorrow and Ill again start updating the blog
by kamatadu, Jun 13, 2022, 11:24 AM
No bengali plx me no understand single word and me too lazy to use gugul translat
by BVKRB-, Jun 7, 2022, 8:55 AM
Amar dadu bhola dadu bhalo dadu noye.
Kamatadu orz
by Bratin_Dasgupta, May 29, 2022, 6:19 AM
hello sar
by anurag27826, May 26, 2022, 1:09 PM
ughh, my time management is literally the worst, cant even figure out time for updating blog
by kamatadu, May 25, 2022, 6:22 PM
ok saar enuf orsing
by BVKRB-, Mar 4, 2022, 5:39 PM
hello sar, plis domt depress by showing ur pure ors skills here pls
by kamatadu, Feb 27, 2022, 3:58 PM
"Hello World?"
by BVKRB-, Feb 27, 2022, 9:57 AM

51 shouts

amar_04 haw so orsmax :omighty:

Set Theory (Pure Basics)

by kamatadu, May 21, 2022, 6:16 PM

3 Comments