LEARN MATH STEP BY STEP THROUGH VERY EASY PROCESS

SOME IMPORTANT THEOREM ON SUBSETS

**Some Theorems On Subset -**

__Theorem.1)__ Every set is a subset of itself.

__Theorem.2)__ The empty set is a subset of
all Sets

**For example, if P = {1, 3, 5, 7}, Q = {1, 3, 5, 7}, then Q ⊆ P and ϕ
⊆ P**

**If, P is a subset of empty
set, if P ⊆ ϕ, then P
= ϕ, **

**P itself is the empty set. Thus ϕ ⊆ ϕ**

__Theorem.3)__ If P ⊆ Q, and Q ⊆ R, then P ⊆ R

__Proof.__
Let, x ∈ P, then since P ⊆
Q, x ∈ Q.

**Now, x ∈ Q**

**=> x ∈ R because Q ⊆ R**

**x ∈ P, and x ∈ R**

**so, P ⊆ R**

__Theorem.4)__ For
any two sets P & Q, if P ⊆ Q, and Q ⊆ P, then P = Q and vice versa

__Proof.-__ Let, x
∈ P, then x ∈ Q since P ⊆ Q …………..(i)

**Also, x ∈ Q, then x ∈ P since Q ⊆
P …………………..(ii)**

**Conversely, let P ⊆ Q, then**

**x ∈ P =>
x ∈ Q, since P = Q,**

**So, P ⊆ Q**

**Similarly, x ∈ Q
=> x ∈ P (since P = Q), **

**So, Q ⊆ P**

**For example, let P = {Richard, Pollard, John}, then all the possible subsets of P are **

**Q = {Richard, Pollard, John}, R
= {Richard, Pollard}, S = {Richard, John}, T = {Pollard, John}, U = {Richard},
V = {Pollard}, W = {John}, X = ϕ**

**The sets Q and X above are
considered as subsets of P, Q ⊆ P and X ⊆
P but sets R, S, T, U, V, and W are considered proper subsets of P, R ⊂ P, S ⊂ P, T ⊂ P, U ⊂ P, V ⊂ P, W ⊂ P.**

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