How have we concluded that X is an inductive set?

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In summary, the proof shows that given any natural numbers $m$ and $n$, if $n$ is an element of $m$, then $n$ is a subset of $m$. This is proven by defining an inductive set $X$ and showing that $X$ contains the necessary elements, namely $\varnothing$ and $n'$ (defined as $n \cup \{n\}$), to be considered an inductive set. This is achieved by showing that $m$ is always a subset of $n$ or $n'$ in both cases in the proof.
  • #1
evinda
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Hello! (Nerd)

I am looking at the proof of the following sentence:

For any natural numbers $m,n$ it holds that:

$$n \in m \rightarrow n \subset m$$

Proof:

We define the set $X=\{ n \in \omega: \forall m (m \in n \rightarrow m \subset n)$ and it suffices to show that $X$ is an inductive set.
Then, $\varnothing \in X$.
Let $n \in X$.
We want to show that $n'=n \cup \{ n \} \in X$.
We pick a $m \in n'=n \cup \{ n \}$.
Then $m \in n$ or $m \in \{ n \}$.
  • If $m \in n$ and since $n \in X$, we have that $m \subset n$
  • If $m \in \{ n \} \rightarrow m=n \rightarrow m \subset n$

So, $n' \in X$ and therefore $X$ is an inductive set.I haven't understood why, having shown that $m \subset n$, we have concluded that $n' \in X$. (Worried)
Could you explain it to me? (Thinking)
 
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  • #2
In order to show that $n' \in X$ don't we have to show that $m \in n' \rightarrow m \subset n'$ ? Or am I wrong? (Thinking)
 
  • #3
In both cases in the proof, we have $m\subseteq n$. This trivially implies $m\subseteq n'$.
 
  • #4
Evgeny.Makarov said:
In both cases in the proof, we have $m\subseteq n$. This trivially implies $m\subseteq n'$.

I see (Nod) Thank you very much! (Happy)
 
  • #5


Sure! Let's break down the proof and see how we can conclude that $X$ is an inductive set.

First, we define the set $X$ as the set of all natural numbers $n$ such that for any natural number $m$, if $m$ is an element of $n$, then $m$ is also a subset of $n$. This definition sets the foundation for our proof.

Next, we show that the empty set, denoted by $\varnothing$, is an element of $X$. This is because the empty set has no elements, so the statement "for any natural number $m$, if $m$ is an element of $\varnothing$, then $m$ is also a subset of $\varnothing$" is automatically true.

Then, we assume that $n$ is an element of $X$. This means that for any natural number $m$, if $m$ is an element of $n$, then $m$ is also a subset of $n$.

Now, we want to show that $n' = n \cup \{n\}$ is also an element of $X$. In order to do this, we need to show that for any natural number $m$, if $m$ is an element of $n'$, then $m$ is also a subset of $n'$.

We have two cases to consider:
1. If $m \in n$, then by our assumption that $n$ is an element of $X$, we know that $m \subset n$.
2. If $m \in \{n\}$, then that means $m$ is equal to $n$. And since we already established in the previous case that $m \subset n$, we can conclude that $m \subset n'$ as well.

Therefore, we have shown that for any natural number $m$, if $m$ is an element of $n'$, then $m$ is also a subset of $n'$. This means that $n'$ satisfies the definition of $X$, and so we can conclude that $n' \in X$.

By repeating this process, we can show that any natural number $n$ is an element of $X$. This is what it means for $X$ to be an inductive set - every natural number is an element of $X$, and for any natural number $n$, $n \in X
 

Related to How have we concluded that X is an inductive set?

1. How do we determine that X is an inductive set?

An inductive set is a set that satisfies the principle of mathematical induction. This means that if the first element of the set is true, and if the truth of any element implies the truth of the next element, then all elements of the set must be true. To determine if a set is inductive, we can use the principle of mathematical induction to show that all elements of the set satisfy this condition.

2. What is the principle of mathematical induction?

The principle of mathematical induction is a proof technique used to show that a statement is true for all elements in a set. It states that if a statement is true for the first element of a set, and if the truth of any element implies the truth of the next element, then the statement must be true for all elements in the set.

3. How is the principle of mathematical induction used to prove that X is an inductive set?

To prove that a set is inductive, we use the principle of mathematical induction to show that if the first element of the set is true, and if the truth of any element implies the truth of the next element, then all elements of the set must be true. This demonstrates that the set satisfies the condition of an inductive set.

4. Can the principle of mathematical induction be used to prove any statement?

The principle of mathematical induction can only be used to prove statements that can be expressed in terms of natural numbers. This is because the principle relies on the fact that natural numbers have a well-defined successor, which is necessary for the proof to work.

5. Are there any other methods for proving that X is an inductive set?

Yes, there are other methods for proving that a set is inductive. One common method is to use structural induction, which is similar to mathematical induction but is used for proving properties of recursively defined objects. Another method is to use transfinite induction, which is used for sets that have an infinite number of elements.

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