MHB Write ⊆ or ∈ in the space provided: {Please Check My Solution}

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The discussion focuses on the correct usage of the symbols ⊆ (subset) and ∈ (element) in set theory. The provided solutions indicate that N is an element of ℚ and P(R), while the empty set ∅ is a subset of Z, and √2 is a subset of R. Participants clarify the definitions, stating that a subset contains all elements of another set, while an element is a member of a set. The conversation also prompts a question about whether N is considered a set or an element. Understanding these concepts is crucial for accurately interpreting set relationships.
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Question:
Write ⊆ or ∈ in the space provided:
N _______ ℚ
N _______ P(R)
∅ _______ Z
√ 2 _______ R

Solution:
N ∈ ℚ
N ∈ P(R)
∅ ⊆ Z
√ 2 ⊆ R
 
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Can you state what $\subseteq$ and $\in$ means?
 
MarkFL said:
Can you state what $\subseteq$ and $\in$ means?
⊆ = Subset
∈ = Element

Definition:

If all the elements of a set is contained in ANOTHER set, then the set whose elements are contained in another set is a subset.
Ex. Set A 's elements= 1,2,3 Set B's elements= 1,2,3,4,5 so... that means all the elements of Set A are in Set B, so A ⊆ B.
 
Good! :D

So, in reference to the first part of the question, is $\mathbb{N}$ a set or an element?
 
I'm taking a look at intuitionistic propositional logic (IPL). Basically it exclude Double Negation Elimination (DNE) from the set of axiom schemas replacing it with Ex falso quodlibet: ⊥ → p for any proposition p (including both atomic and composite propositions). In IPL, for instance, the Law of Excluded Middle (LEM) p ∨ ¬p is no longer a theorem. My question: aside from the logic formal perspective, is IPL supposed to model/address some specific "kind of world" ? Thanks.
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...

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