Hi again!

Ususally I stick to making my notations as simple as possible,but this time it will get a little complex.To make sure,that I minimalize the amount of mistakes and wrong assumptions I make,I'll start with it simple,but the definitions at the end are going to be more complicated.Like always,if you have any questions or see a mistake in my work,then make sure you tell me in the coments.

Let \(\mathbb{N}\) be the set of all natural numbers.

Let \(\mathbb{Q}\) be the set of all rational numbers.

It all starts from here.

It's not very hard to show that 

\(\mathbb{N} \subseteq \mathbb{Q} \Leftrightarrow \mathbb{Q} \supseteq \mathbb{N} \land \mathbb{N} \subset \mathbb{Q} \Leftrightarrow \mathbb{Q} \supset \mathbb{N}\)

Let \(n\) be an element of \(\mathbb{N}\).

\(n \in \mathbb{N}\)

Let \(j\) be an element of \(\mathbb{Q}\) but not of \(\mathbb{N}\).

\(j \in \mathbb{Q} \land j \notin \mathbb{N}\)

Let \(S1\) and \(S2\) be two distinct sets.

\(\forall S1,S2,k:(S1 \subseteq S2 \land k \in S1) \Rightarrow k \in S2\)

Let i be an element of either \(\mathbb{N}\) or \(\mathbb{Q}\) but it is not known which.

\(i \in (\mathbb{N} \lor \mathbb{Q})\) 

Let \(x_k \in S1\) be the k-th element in the set \(S1\).This is important because it will come in handy later on.

\(\forall i,k:(i < k \land i \in \mathbb{N} \land k \in \mathbb{N} \land \mathbb{N} \subseteq \mathbb{Q}) \Rightarrow i \in \mathbb{Q} \land k \in \mathbb{Q} \land x_i \in (\mathbb{N} \lor \mathbb{Q}) < x_k \in (\mathbb{N} \lor \mathbb{Q})\)

Now we've proven that all:

\(\forall n,j,i:(n \in \mathbb{N} \land j \in \mathbb{Q} \land j \notin \mathbb{N} \land i \in (\mathbb{Q} \lor \mathbb{N}) \land \mathbb{N} \subseteq \mathbb{Q}) \Rightarrow n \in \mathbb{Q} \land j \in \mathbb{Q} \land i \in \mathbb{Q}\)

Next,we keep the definitions for \(n,j\) and \(i\) and we prove something else.

\(\forall S1,S2:(S1 \subseteq S2) \Rightarrow S1 \cap S2 = S2 \vdash n \in (\mathbb{Q} \cap \mathbb{N}) \land j \in (\mathbb{Q} \backslash \mathbb{N}) \land j \notin (\mathbb{Q} \cap \mathbb{N}) \land i \in (\mathbb{Q} \cap \mathbb{N} \lor \mathbb{Q} \backslash \mathbb{N})\)

Now we define a new set \(P\) as:

\(P \subseteq \mathbb{Q} \land P \cap \mathbb{N} = \emptyset \Rightarrow P = \mathbb{Q} \backslash \mathbb{N}\)

\(\forall S1,S2:(S1 \subset S2) \Rightarrow \neg (S1 \subseteq (S2 \backslash S1))\)

\(\forall S1,S2:(S1 \subset S2) \Rightarrow (S2 \backslash S1) \supset S2 \land \neg (S1 \cap S2 \subseteq (S2 \backslash S1))\)

\(j \in \mathbb{Q} \land j \notin \mathbb{N} \Rightarrow j \in \mathbb{Q} \backslash \mathbb{N} \Rightarrow j \in P\)

Let \(\mathbb{Z}\) be the set of all integers.

\(\forall S1:(S1 \subseteq \mathbb{Z} \cap \mathbb{Q}) \Rightarrow S1 \subset \mathbb{Z} \cup \mathbb{Q} \Rightarrow S1 \subset \mathbb{Z} \land S1 \subset \mathbb{Q}\)

We define \(S\) to be equivalent to some set \(\{n_1,n_2,....,n_j\}\) such that \(\forall k \in \mathbb{N} \Rightarrow n_k \in \mathbb{N}\)

We define \(T\) such that \([S]T\) adds 1 to all elements of the string \(S\) and we define \(T-\) such that \([S]T-\) remeoves the first element of the string and adds one to the last,then shifts to \(T\) and \(S-\) such that \([S-]\#\) remeves all elements of of the string \(S\) that are homeomorphically embeddable to any other part of the string and it shifts back to \(S\) normally,adding an extra \(T\) if there is no existing one,otherwise if there is a \(T\) already existing it shifts it to \(T-\).Note:\(S-\) has to be done before any other operation.

We define \(V_k\) such that \([S]TV_{k+1} = [S-]TV_k\) and \(V_0\) is removed.

Now you may be asking "What was this function for?" and the reason I made it was for it to be a simple combinatorical function which will be the base for what I will define later.

We define the function \(pk(\) where \(n\) is the number and \(abc....\) are the digits after the decimal to be equal to \([....,c,b,a]TV_n\).It should be clear now that for \(pk(n)\) to give some finite value,\(n\) must not be irrational.

\(\forall n,k:(pk(n) = k \land k \in \mathbb{N}) \Rightarrow n \in \mathbb{Q}\)

We define \(\varphi\) to be some element of \(\mathbb{N}\) such that there is no rational number \(\psi\) that when inserted in \(pk()\) will give out \(\varphi\) as an output.

\(\not\exists \psi(pk(\psi)=\varphi) \Leftrightarrow \psi \in P\)

Now there are many \(\varphi\) that fit this definition,perhaps infinitely,so we define \(\varphi[\alpha]\) as the \(\alpha\)-th \(\varphi\) fitting these definitions.

\(\varphi[k] = x_k \in \mathbb{N}(\not\exists \psi(pk(\psi)=\varphi) \Leftrightarrow \psi \in P)\)

Let \(*\) be a Kleene star,indicating that a set \(S1\),when turned into \(S1^*\) is not a set of subsets but an ordered tuple of sets with the same properties as those in the original set.

Let \(\hat{\bigcup}\) be a special type of universal set,which is a collection of all sets and elements which fit the surtain properties we search for.For example \(n = \hat{\bigcup}a(a \in \mathbb{N} \land a>5)\) then \(n = \{6,7,8,...\}\).

\(\hat{\bigcup}S1\) also describes all the elements of the set \(S1\) but when some condition is applied,only the elements that satisfy that condition remain.

Then let \(\subset^*\) be a special type of subsetting such that it describes all elements in the first set mapped in correspondence to all elements of the second step map,with the correspondence of mapping the two sets being the sets transformed into ordered tuples.The same goes for \(\subseteq^*,\supset^*,\supseteq^*\).Formally,we can use the new defined operators in this definition:\(S1 \subseteq^* S2 \Leftrightarrow \hat{\bigcup}S1^* \mapsto \hat{\bigcup}S2^*\) 

Similarly,\(S1 \subset^* S2\) is the same but \(S1^* \neq S2^* \Rightarrow S1 \neq S2\).

Let \(R\) be the set of all \(k\) such that:

\(k \in \mathbb{N} \land k \mapsto x_k \in \hat{\bigcup}P^*\)

Using our new language,\(R = \hat{\bigcup}k(k \in \mathbb{N} \land k \mapsto x_k \in \hat{\bigcup}P^*)\)

And let \(R\) be short for \(R_0\) and:

\(R_{n+1} = \hat{\bigcup}k(k \succ \hat{\bigcup}\varphi \in \varphi[pk(\alpha)]\forall \alpha \in P \land \alpha \notin (\mathbb{N} \backslash R_n)^*\)

Let \(\succ\) indicate sequence domination .

Let \(A\) be a tuple \((a_1,a_2,.....,a_j)\) where \(a_i\) is an element of \(R_i\) such that:

\(\forall j:(j \in \mathbb{N})\exists !b(a_j \prec b \prec a_{2j}) \Leftrightarrow b \in P \land b \notin A \Rightarrow \varphi[b] > \varphi[a_j]\)

Let \(O\) be the set of all b for which this would work.

\(O = \hat{\bigcup}b(a_j \prec b \prec a_{2j} \forall j \in \mathbb{N} \Leftrightarrow b \in P \land b \notin A \Rightarrow \varphi[b] > \varphi[a_j])\)

\(\forall S1:(S1 \subseteq R_k)\forall k \in \mathbb{N}\exists S2 \subseteq O(S2 \succ S1)\exists S3 \subseteq R_i(S1 \subseteq^* S3 \land S3 \prec S2)\)

With these definitions \(x_n \in O^*\) is the \(n\)-th element of the ordered tuple of the set \(O\).

Let \(\top\) mean a statement that is unconditionally true and \(\bot\) be a statement that is unconditionally false and let \(\top(\alpha)\) mean that a very specific \(\alpha\) is proven to be unconditionally true and similarly,let \(\bot(\alpha)\) be a very specific \(\alpha\) is proven to be unconditionally false.

Now let the definition of a new set \(F\) to be:

\(F = \hat{\bigcup}k(k = \varphi[\psi] \Leftrightarrow pk(n) = i \land \psi = x_i \in \hat{\bigcup}\alpha(\alpha \not\prec j)\forall j \in S1 \land S1 \subset^* O)\)

We are almost finished.Just one more painful definition.

\(F(n) = x_n \in \hat{\bigcup}\psi(\varphi[\psi] \succ \hat{\bigcup}\varphi[\gamma]\forall \gamma:(\gamma < \psi) \lor (\psi \in S1 \land \gamma \notin S1 \land S1 \subset^* F) \land \psi \succ \hat{\bigcup}\mu(\\ \mu \in (\top(\mu) \vdash (\bot(\gamma) \land \bot(\hat{\bigcup}\neg \varphi(\varphi \notin \varphi[\alpha]\forall \alpha < \mu))))))\land \psi \in F)\)

Whoa,that is long!

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