# What is the least upper bound ordinal of my linear n-symbol partition ordinal (ordinal that contain all finite string of finite different symbol)?

Note: The "partition" here isn't relate to the partition at all.

Note: For the detail of "ordinal that contain all finite string of finite different symbols", see the "Edit" section at the end of this post.

I start with the idea that if I have more symbols, I can represent more distinct things. And if I have more different ways to write symbols, I can represent more distinct things as well.

So, I start at the simplest form. I have invented what I call the linear n-symbol partition ordinal. I call it "linear" because it is a string of symbols in single line, without subscripts or superscripts. I will talk about what I mean by "partition" later.

Let say the first symbol is $$0$$.

With only one symbol, I can represent any finite ordinal in my ordinal system. By definition, any string of just 0's has the value of the number of zeros in the string:

\begin{align} \text{(My Notation)} &=> \text{(Conventional Notation)} \\ 0 &=> 0 \\ 00 &=> 1 \\ 000 &=> 2 \\ 0000 &=> 3 \\ &... \\ \end{align}

However, the single symbol $$0$$ can't represent ω, so the least upper bound ordinal of this 1-symbol partition ordinal is ω.

This where the second symbol comes in.

$$1$$ is set of all ordinals that can be represent with finite number of $$0$$ symbols. If zeros come after, they are added in the standard ordinal-arithmetic sense:

\begin{align} \text{(My Notation)} &=> \text{(Conventional Notation)} \\ 1 &=> ω \\ 10 &=> ω+1 \\ 100 &=> ω+2 \\ &... \\ \end{align}

When we have two or more symbols, we start to have partitions. The second symbol acts as wall that I call a partition of the first symbol.

Where zeros come before the 1, each zero indicates adding the set to itself one time. Again, comparing to conventional notation:

\begin{align} \text{(My notation)} &=> \text{(Conventional notation)} \\ 01 = 1+1 &=> ω+ω = ω*2 \\ 010 &=> ω*2+1 \\ 0100 &=> ω*2+2 \\ 001 = 1+1+1 &=> ω*3 \\ 0001 &=> ω*4 \\ 00001 &=> ω*5 \end{align}

And you can see, the $$0$$ in front of $$1$$ can be added infinitely and the least upper bound is ω*ω. To extend this, I add one more symbol $$1$$

$$101$$ is the product of $$1$$ and $$1$$ in ordinal arithmetic. Each additional zero between the ones add an $$ω$$ to the value:

\begin{align} \text{(My Notation)} &=> \text{(Conventional Notation)} \\ 101 = 1 * 1 &=> ω*ω \\ 1010 &=> ω*ω + 1 \\ 1001 &=> ω*ω + ω \\ 10001 &=> ω*ω + ω*2 \\ 100001 &=> ω*ω + ω*3 \\ \end{align}

Similar to adding zeros in front of a single $$1$$, each $$0$$ indicates adding the set to itself one time. Again, comparing to conventional notation:

\begin{align} \text{(My Notation)} &=> \text{(Conventional Notation)} \\ 0101 = 101 + 101 &=> ω*ω + ω*ω = ω*ω*2 \\ 00101 &=> ω*ω*3 \\ \end{align}

And each additional $$10$$ group prepended multiplies by $$ω$$:

\begin{align} \text{(My Notation)} &=> \text{(Conventional Notation)} \\ 10101 = 1 * 1 * 1 &=> ω*ω*ω \\ 1010101 = 1 * 1 * 1 * 1 &=> ω*ω*ω*ω \end{align}

As you can see, $$1$$ with a $$0$$ to separate can be added infinitely many times. To extend this, let use $$11$$

$$11$$ is set of all ordinals that

• Can be represent with finite $$1$$ symbol.
• Have finite $$0$$ symbols.
• Don't have any $$1$$'s next to each other.

At this point, I will call $$1$$, $$11$$, and other string without $$0$$ symbol as partition instead. The partition can't be next to each other without $$0$$ in between them.

That mean definition of $$11$$ is set of all ordinals that

• Can be represent with finite $$1$$ partition.
• Have finite $$0$$ symbol.

I don't have to say "no any $$1$$ next to each other" because partition can't be next to each other without $$0$$ in between them. The $$1$$ next to each other will be other partition.

$$1101$$ is set of all ordinals that

• That is the member of the set $$11$$.
• And some ordinals that isn't the member of the set $$11$$ but those ordinals have the following property
• Can be represent 1 "11" partition
• Have no "1" partition in front of the rightmost "11" partition.
• Have no "0" symbol in front of the rightmost "11" partition.
• Have no "1" partition after the rightmost "11" partition.
• Have finite "0" symbol after the rightmost "11" partition.

"110101" is set of all ordinals that

• That is the member of the set "11".
• And some ordinals that isn't the member of the set "11" but those ordinals have the following property
• Can be represent 1 "11" partition
• Have no "1" partition in front of the rightmost "11" partition.
• Have no "0" symbol in front of the rightmost "11" partition.
• Have 1 or less "1" partition after the rightmost "11" partition.
• Have finite "0" symbol after the rightmost "11" partition.

\begin{align} 11 &= 1 \cup \{1,...,101,...,10101,...,1010101,...\} \\ 1101 &= 11 \cup \{11,110,1100,11000,...\} \\ 110101 &= 1101 \cup \{1101,...,11001,...,110001,...\} \\ \end{align}

$$011$$ is set of all ordinals that

• That is the member of the set $$11$$.
• And some ordinals that isn't the member of the set "11" but those ordinals have the following property
• Can be represent 1 "11" partition
• Have no "1" partition in front of the rightmost "11" partition.
• Have no "0" symbol in front of rightmost "11" partition.
• Have finite "1" partition after the rightmost "11" partition.
• Have finite "0" symbol after the rightmost "11" partition.

"1011" is set of all ordinals that

• That is the member of the set "11".
• And some ordinals that isn't the member of the set "11" but those ordinals have the following property
• Can be represent 1 "11" partition
• Have no "1" partition in front of the rightmost "11" partition.
• Have finite "1" partition after the rightmost "11" partition.
• Have finite "0" symbol.

"101011" is set of all ordinals that

• That is the member of the set "1011".
• And some ordinals that isn't the member of the set "1011" but those ordinals have the following property
• Can be represent 1 "11" partition
• Have 1 "1" partition in front of the rightmost "11" partition.
• Have finite "1" partition after the rightmost "11" partition.
• Have finite "0" symbol.

"10101011" is set of all ordinals that

• That is the member of the set "101011".
• And some ordinals that isn't the member of the set "101011" but those ordinals have the following property
• Can be represent 1 "11" partition
• Have 2 "1" partition in front of the rightmost "11" partition.
• Have finite "1" partition after the rightmost "11" partition.
• Have finite "0" symbol.

\begin{align} 011 &= 11 \cup \{11,...,1101,...,110101,...,11010101,...\} \\ 1011 &= 11 \cup \{11,...,1101,...,110101,...,011,...,...,0011,...,...,00011,...\} \\ 101011 &= 1011 \cup \{1011,...,10011,...,100011,...,01011,...,...,001011,...,...,0001011,...\} \\ 10101011 &= 101011 \cup \{101011,...,1001011,...,10001011,...,0101011,...,...,00101011,...,...,000101011,...\} \\ \end{align}

"11011" is set of all ordinals that

• Can be represent 1 "11" partition
• Have finite "1" partition.
• Have finite "0" symbol.

"110011" is set of all ordinals that

• That is the member of the set "11011".
• And some ordinals that isn't the member of the set "11011" but those ordinals have the following property
• Can be represent 2 "11" partition
• Have no "1" partition in front of the leftmost "11" partition.
• Have no "0" symbol in front of the leftmost "11" partition.
• Have no "1" partition between each "11" partition.
• Have 1 or less "0" symbol between each "11" partition.
• Have finite "1" partition after the rightmost "11" partition.
• Have finite "0" symbol after the rightmost "11" partition.

"1101011" is set of all ordinals that

• That is the member of the set "11011".
• And some ordinals that isn't the member of the set "11011" but those ordinals have the following property
• Can be represent 2 "11" partition
• Have no "1" partition in front of the leftmost "11" partition.
• Have no "0" symbol in front of the leftmost "11" partition.
• Have no "1" partition between each "11" partition.
• Have finite "0" symbol between each "11" partition.
• Have finite "1" partition after the rightmost "11" partition.
• Have finite "0" symbol after the rightmost "11" partition.

"110101011" is set of all ordinals that

• That is the member of the set "1101011".
• And some ordinals that isn't the member of the set "1101011" but those ordinals have the following property
• Can be represent 2 "11" partition
• Have no "1" partition in front of the leftmost "11" partition.
• Have no "0" symbol in front of the leftmost "11" partition.
• Have 1 or less "1" partition between each "11" partition.
• Have finite "0" symbol between each "11" partition.
• Have finite "1" partition after the rightmost "11" partition.
• Have finite "0" symbol after the rightmost "11" partition.

\begin{align} 11011 &= 11 \cup \{11,...,1011,...,101011,...,10101011,...\} \\ 110011 &= 11011 \cup \{11011,...,1101101,...,110110101,...,11011010101,...\} \\ 1101011 &= 11011 \cup \{11011,...,1101101,...,110110101,...,110011,...,...,1100011,...,...,11000011,...\} \\ 110101011 &= 1101011 \cup \{1101011,...,11010011,...,110100011,...,11001011,...,...,110001011,...,...,1100001011,...\} \\ \end{align}

"011011" is set of all ordinals that

• That is the member of the set "11011".
• And some ordinals that isn't the member of the set "11011" but those ordinals have the following property
• Can be represent 2 "11" partition
• Have no "1" partition in front of the leftmost "11" partition.
• Have no "0" symbol in front of the leftmost "11" partition.
• Have finite "1" partition after the leftmost "11" partition.
• Have finite "0" symbol after the leftmost "11" partition.

"1011011" is set of all ordinals that

• That is the member of the set "11011".
• And some ordinals that isn't the member of the set "11011" but those ordinals have the following property
• Can be represent 2 "11" partition
• Have no "1" partition in front of the leftmost "11" partition.
• Have finite "0" symbol in front of the leftmost "11" partition.
• Have finite "1" partition after the leftmost "11" partition.
• Have finite "0" symbol after the leftmost "11" partition.

"11011011" is set of all ordinals that

• Can be represent 2 "11" partition
• Have finite "1" partition.
• Have finite "0" symbol.

"11011011011" is set of all ordinals that

• Can be represent 3 "11" partition
• Have finite "1" partition.
• Have finite "0" symbol.

\begin{align} 011011 &= 11011 \cup \{11011,...,110011,...,1100011,...,1101011,...,...,110101011,...,...,110101011,...\} \\ 1011011 &= 11011 \cup \{11011,...,1101011,...,110101011,...,011011,...,...,0011011,...,...,00011011,...\} \\ 11011011 &= 11011 \cup \{11011,...,011011,...,0011011,...,1011011,...,...,101011011,...,...,10101011011,...\} \\ 11011011011 &= 11011011 \cup \{11011011,...,011011011,...,0011011011,...,1011011011,...,...,101011011011,...,...,10101011011011,...\} \\ \end{align}

Compare to mathematical ordinal number

\begin{align} \text{(My Ordinal)} &=> \text{(Math Ordinal)} \\ 11 &=> ω^ω \\ 1101 &=> ω^ω + ω \\ 110101 &=> ω^ω + ω*2 \\ 011 = 11 + 11 &=> ω^ω + ω^ω = (ω^ω)*2 \\ 1011 &=> (ω^ω)*ω = ω^{ω+1} \\ 10011 &=> (ω^ω)*ω+ω^ω \\ 01011 &=> (ω^ω)*ω+(ω^ω)*ω = (ω^ω)*ω*2 \\ 101011 &=> (ω^ω)*ω*ω = ω^{ω+2} \\ 10101011 &=> ω^{ω+3} \\ 11011 = 11 * 11 &=> ω^{ω+ω} \\ 110011 = 11011 + 11 &=> ω^{ω+ω}+ω^ω \\ 1101011 = 11011 + 1011 &=> ω^{ω+ω}+ω^{ω+1} \\ 011011 = 11011 + 11011 &=> ω^{ω+ω}*2 \\ 1011011 &=> ω^{ω+ω}*ω = ω^{ω+ω+1} \\ 101011011 &=> ω^{ω+ω}*ω*ω = ω^{ω+ω+2} \\ 11011011 &=> ω^{ω+ω}*ω^ω = ω^{ω+ω+ω} \\ 11011011011 &=> ω^{ω+ω}*ω^ω*ω^ω = ω^{ω+ω+ω+ω} \end{align}

At this point, I see some pattern. It isn't proved to be correct but it is never wrong.

Inductive reasoning: If $$A$$ is the ordinal that its first character isn't $$0$$, $$0A = A + A$$. For example, $$011011 = 11011 + 11011$$.

Inductive reasoning: If $$A$$ is 1-level partition ordinal (string without any symbol that is lower than "1"), A0A = A * A. For example, $$11011 = 11 * 11$$.

"higher" and "lower" symbol mean that, for example,

• $$0$$ is lower than $$1$$
• $$1$$ is higher than $$0$$
• Any symbol that is lower than "4" are "0","1","2" and"3"
• Any symbol that is higher than "4" are "5","6","7",...

"n-level partition ordinal" are ordinal that don't have symbol that is lower than "n", for example, 4-level partition ordinal is ordinal that don't have symbol that is lower than "4".

As you can see, $$11$$ can be add infinitely and the least upper bound is $$ω^{ω^ω}$$. To extend this, let use $$111$$

"111" is set of all ordinals that

• Have finite "11" partition
• Have finite "1" partition.
• Have finite "0" symbol.

"0111" is set of all ordinals that

• That is the member of the set "111".
• And some ordinals that isn't the member of the set "111" but those ordinals have the following property
• Can be represent 1 "111" partition
• Have no "11" partition in front of the rightmost "111" partition.
• Have no "1" partition in front of the rightmost "111" partition.
• Have no "0" symbol in front of the rightmost "111" partition.
• Have finite "11" partition after the rightmost "111" partition.
• Have finite "1" partition after the rightmost "111" partition.
• Have finite "0" symbol after the rightmost "111" partition.

"10111" is set of all ordinals that

• That is the member of the set "111".
• And some ordinals that isn't the member of the set "111" but those ordinals have the following property
• Can be represent 1 "111" partition
• Have no "11" partition in front of the rightmost "111" partition.
• Have no "1" partition in front of the rightmost "111" partition.
• Have finite "0" symbol in front of the rightmost "111" partition.
• Have finite "11" partition after the rightmost "111" partition.
• Have finite "1" partition after the rightmost "111" partition.
• Have finite "0" symbol after the rightmost "111" partition.

"110111" is set of all ordinals that

• That is the member of the set "111".
• And some ordinals that isn't the member of the set "111" but those ordinals have the following property
• Can be represent 1 "111" partition
• Have no "11" partition in front of the rightmost "111" partition.
• Have finite "1" partition in front of the rightmost "111" partition.
• Have finite "0" symbol in front of the rightmost "111" partition.
• Have finite "11" partition after the rightmost "111" partition.
• Have finite "1" partition after the rightmost "111" partition.
• Have finite "0" symbol after the rightmost "111" partition.

"1110111" is set of all ordinals that

• Can be represent 1 "111" partition
• Have finite "11" partition
• Have finite "1" partition.
• Have finite "0" symbol.

"11101110111" is set of all ordinals that

• Can be represent 2 "111" partition
• Have finite "11" partition
• Have finite "1" partition.
• Have finite "0" symbol.

"1111" is set of all ordinals that

• Have finite "111" partition
• Have finite "11" partition
• Have finite "1" partition.
• Have finite "0" symbol.

"11111" is set of all ordinals that

• Have finite "1111" partition
• Have finite "111" partition
• Have finite "11" partition
• Have finite "1" partition.
• Have finite "0" symbol.

\begin{align} 111 &= 11 \cup \{11,...,11011,...,11011011,...,11011011011,...\} \\ 0111 &= 111 \cup \{111011,...,111011011,...,111011011011,...,111011011011011,...\} \\ 10111 &= 111 \cup \{111,...,0111,...,00111,...,000111,...\} \\ 110111 &= 111 \cup \{111,...,10111,...,1010111,...,101010111,...\} \\ 1110111 &= 111 \cup \{111,...,110111,...,110110111,...,110110110111,...\} \\ 11101110111 &= 1110111 \cup \{1110111,...,1101110111,...,1101101110111,...,1101101101110111,...\} \\ 1111 &= 111 \cup \{111,...,1110111,...,11101110111,...,111011101110111,...\} \\ 11111 &= 1111 \cup \{1111,...,111101111,...,11110111101111,...,1111011110111101111,...\} \\ \end{align}

I will stop compare to mathematical ordinal number here because it will make the post too long. But I will explain how my encoding method work instead.

The least upper bound ordinal of 2-symbol partition ordinal is $$ω^{ω^ω}$$. To extend this, I have to add one more symbol "2".

The "2" symbol is partition of "1" and "0".

"2" is set of all ordinals that

• Have finite "1" symbol.
• Have finite "0" symbol.

That is the simplest definition of "2" but might cause some confusing. So, here is the more precise definition of "2"

"2" is set of all ordinals that

• Can be represent finite number of partitions which have finite "1" symbol.
• Have finite "0" symbol.

"201" is set of all ordinals that

• That is the member of the set "2".
• And some ordinals that isn't the member of the set "2" but those ordinals have the following property
• Can be represent 1 "2" partition
• Have no partitions which have 1 or more "1" symbol.
• Have no "0" symbol in front of the rightmost "2" partition.
• Have finite "0" symbol after the rightmost "2" partition.

"2011" is set of all ordinals that

• That is the member of the set "2".
• And some ordinals that isn't the member of the set "2" but those ordinals have the following property
• Can be represent 1 "2" partition
• Have no partitions which have 1 or more "1" symbol in front of the rightmost "2" partition.
• Have no "0" symbol in front of the rightmost "2" partition.
• Have no partitions which have 2 or more "1" symbol after the rightmost "2" partition.
• Have finite number of partitions which have 1 or less "1" symbol after the rightmost "2" partition.
• Have finite "0" symbol after the rightmost "2" partition.

"02" is set of all ordinals that

• That is the member of the set "2".
• And some ordinals that isn't the member of the set "2" but those ordinals have the following property
• Can be represent 1 "2" partition
• Have no partitions which have 1 or more "1" symbol in front of the rightmost "2" partition.
• Have no "0" symbol in front of the rightmost "2" partition.
• Have finite number of partitions which have finite "1" symbol after the rightmost "2" partition.
• Have finite "0" symbol after the rightmost "2" partition.

"102" is set of all ordinals that

• That is the member of the set "2".
• And some ordinals that isn't the member of the set "2" but those ordinals have the following property
• Can be represent 1 "2" partition
• Have no partitions which have 1 or more "1" symbol in front of the rightmost "2" partition.
• Have finite "0" symbol in front of the rightmost "2" partition.
• Have finite number of partitions which have finite "1" symbol after the rightmost "2" partition.
• Have finite "0" symbol after the rightmost "2" partition.

"1102" is set of all ordinals that

• That is the member of the set "2".
• And some ordinals that isn't the member of the set "2" but those ordinals have the following property
• Can be represent 1 "2" partition
• Have no partitions which have 2 or more "1" symbol in front of the rightmost "2" partition.
• Have finite number of partitions which have 1 or less "1" symbol in front of the rightmost "2" partition.
• Have finite "0" symbol in front of the rightmost "2" partition.
• Have finite number of partitions which have finite "1" symbol after the rightmost "2" partition.
• Have finite "0" symbol after the rightmost "2" partition.

"202" is set of all ordinals that

• Can be represent 1 or less "2" partition.
• Have finite number of partitions which have finite "1" symbol.
• Have finite "0" symbol.

\begin{align} 2 &= 2 \cup \{1,...,11,...,111,...,1111,...\} \\ 201 &= 2 \cup \{2,20,200,2000,...\} \\ 2011 &= 2 \cup \{2,20,200,...,201,...,20101,...,2010101,...\} \\ 02 &= 2 \cup \{2,...,201,...,20101,...,2011,...,...,20111,...,...,201111,...\} \\ 102 &= 2 \cup \{2,...,2011,...,20111,...,02,...,...,002,...,...,0002,...\} \\ 1102 &= 2 \cup \{2,...,02,...,002,...,102,...,...,10102,...,...,1010102,...\} \\ 202 &= 2 \cup \{102,...,1102,...,11102,...,111102,...\} \\ \end{align}

I will also go faster and skip all non 1-level partition ordinal and focus only on 1-level partition ordinal. Once we have 3 symbol, if I keep explain the ordinal that have "0" symbol. It will make the post too long. However, I think you can see the pattern.

In 1-level partition, "1" is the lowest symbol while "2" and higher symbol are partitions. Like the 0-level (or the lowest level) when "0" is the lowest symbol while "1" and higher symbol are partitions.

As you can see, when we have many but finite $$11$$ partition, the set that contain all of them is $$111$$. And when we have many but finite $$111$$ or less partition, the set that contain all of them is $$1111$$. So, when we have many but finite $$2$$ or less partition, the set that contain all of them is $$21$$.

"A" or less partition mean "A" and all 1-level partition ordinal that is member of set "A"

So, the set $$21$$ is set of all ordinals that

• Can be represented by a finite number of 1-level partitions which have the following property
• That is the member of the set $$2$$.
• And some 1-level partitions that isn't the member of the set "2" but those ordinals have the following property
• Have 1 "2" partition.
• Have no "1" symbol.
• Have finite "0" symbol.

\begin{align} 21 &=> 2 \cup \{2,...,102,...,1102,...,202,...,...,20202,...,...,2020202,...\} \\ \end{align}

And we can make the set $$211$$,$$2111$$,... and so on with the same method that we make $$21$$ from $$2$$. The set that contains them all is $$21$$.

The set $$12$$ is set of all ordinals that

• Can be represent finite number of 1-level partitions which have the following property
• That is the member of the set "2".
• And some 1-level partitions that isn't the member of the set "2" but those ordinals have the following property
• Have 1 "2" partition.
• Have no "1" symbol in front of the rightmost "2" partition.
• Have finite "1" symbol after the rightmost "2" partition.
• Have finite "0" symbol.

\begin{align} 12 &=> 2 \cup \{2,...,21,...,211,...,2111,...\} \\ \end{align}

And again, we can make the set "121","1211",... and so on with that same method. And set that contain them all is "112".

So, we can make set "1112", "11112",... and so on with that method. The set that contain them all is "212".

The set "212" is set of all ordinals that

• Can be represent finite number of 1-level partitions which have the following property
• That is the member of the set "2".
• And some 1-level partitions that isn't the member of the set "2" but those ordinals have the following property
• Have 1 "2" symbol.
• Have finite "1" symbol.
• Have finite "0" symbol.

\begin{align} 212 &=> 2 \cup \{2,...,21,...,211,...,12,...,...112,...,...,1112,...\} \\ \end{align}

If you can remember $$101$$, you will found that $$212$$ in the 1-level is similar to $$101$$ in the 0-level.

Yes, in 1-level, "1" is like "0" in the 0-level and "2" is like "1" in 0-level. The only different between them is that adding "1" on the rightmost take many more step than adding "0" the rightmost. However, we can use adapt the method how we define "1","11",... partitions in 0-level to define "2","22",... partitions in 0-level.

Note : Like "0" in "101", the "1" in "212" is needed to separate two "2" and doesn't have any value at all.

So, we can make the set $$21212$$,$$2121212$$,... and so on. And set that contains them all is $$22$$.

We can make the set $$22122122$$,$$22122122122$$,... and so on. And set that contain them all is $$222$$.

Then we can make the set $$2222$$,$$22222$$,... and so on. To extend this, I have to add one more symbol $$3$$. The set that contain them all is $$3$$. The 3-symbol will have least upper bound somewhere at the "3".

Back to the question. What is the least upper bound ordinal of my linear n-symbol partition ordinal?

As you can see, even though I added more symbols there must still be a least upper bound somewhere.

The question is, what is the least upper bound that my ordinal can't reach no matter how many finite number of symbols I have?

I can work step by step but as there are more symbol, the step by step is longer. At some point, it will take too long to do so.

I suspect that least upper bound of my linear n-symbol partition ordinal is $$ε_0$$ but don't know how to prove/disprove it.

Edit: After I fix this post, I realize that the set $$n$$ will contain all finite string that made of finite $$n-1$$ symbol or less. For example,

• The set $$2$$ contain all finite string that made of finite $$0$$ and $$1$$ symbol.
• The set $$3$$ contain all finite string that made of finite $$0$$, $$1$$ and $$2$$ symbols.

That mean the least upper bound ordinal of my linear $$n$$-symbol partition ordinal will contain all finite strings that can represent with finite different symbols.

• $\epsilon_0$ is correct. You need to prove that you can produce any finite power tower of $\omega$'s (easy), and that every number you produce is less than some finite power tower of $\omega$'s. I think a good first step in proving this would be to write down some structural formulae that define your encoding method, instead of a large list of examples. Commented Jul 9, 2023 at 17:23
• @MikeEarnest Could you tell me or give me an example how to write down the structural formulae mathematically ? I think I know its structural formulae but I don't know how to write it down mathematically. Commented Jul 9, 2023 at 17:50
• If you could describe, in plain English, the process that decodes a number-sequence to its corresponding ordinal, then that would be a big help. Basically, when I said "formula" in my first comment, I should have said that English or pseudo-code wold be OK. Commented Jul 9, 2023 at 18:34
• @MikeEarnest Ok, I will try. However, after I think again, I think I might not know the exact structural formulae. I know how to work up to any number-sequence (It may take very long step but I know the exactly way.) but the isn't equal to the structural formulae. There are many formula that I know base on the observation that I say "Inductive reasoning". I know it doesn't mean that formula is proved to be correct but I don't see any example that make the observation wrong. Commented Jul 9, 2023 at 19:35
• @MikeEarnest I try to fix it. I know it is very long but I don't know how to explain it shorter. It is just too different from how normal ordinal work. And I have to explain everything. If you can make it shorter but still contain the same information, I will be great. Commented Jul 14, 2023 at 20:26