But I thought that you already agreed that there is such a place, if not explicitly, then at least implicitly (: In the case of ((1, 2, 3, \dotso, a)), that place is the one occupied by (a). Didn’t you agree that the index of that place is (infA + 1)? If so, that’s a “point at (infA)”.
Let me restate the entire logic:
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“Point” is another word for “position” or “place”
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Thus, “point at infinity” is another expression for “place at infinity”
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“Place at X” is another expression for “place whose index is X”
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Thus, “place at infinity” is another expression for “place whose index is infinity”
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(infA) is an instance of infinity
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Thus, “place whose index is (infA)” is also “place whose index is infinity”
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The index of (a) in ((1, 2, 3, \dotso, a)) is (infA + 1)
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Thus, (a) occupies “place whose index is (infA + 1)” which means “point at (infA + 1)” which means “point at infinity”
Please let me know what you disagree with.
I don’t think that “quality” is an appropriate way to describe what (infA) is. You can’t even say that (infA) specifies that a set has no end because sets have no notion of “end”. (“Endless set” is an instance of figurative speech.) What (infA) does is it specifies how many elements there are in a set. Specifically, what it does is it states that the number of elements in a set is equal to the number of natural numbers.
It is an index of the place (which is a point) occupied by (a) in the case of ((2, 3, 4, \dotso, a)).
I am not sure I understand what you’re saying here. Are you saying that there is no “end” to the sequence that is ((1, 2, 3, \dotso, a))? But there is. Remember how we defined the word “end”? It refers to a place that comes after all other places i.e. to a place with the highest index. And that place in the case of ((1, 2, 3, \dotso, a)) is the one occupied by (a).
If this is true then ((1, 2, 3, \dotso, a)) is an invalid sequence. Do you agree with that?
I didn’t add (a) to a new sequence. That’s NOT the operation I performed. I added it to THE SAME sequence and I added it AFTER all of its elements. You are insisting that I performed an operation that I did not actually perform. (It’s akin to saying I multiplied (1) by (0) when in fact I divided it by (0).) You may want to argue that I’m performing an impossible operation instead (which is false but which is at least based on what I really did rather than something that I didn’t do.)
And note that when we say that a sequence has no end, we’re merely saying that it has no place with the highest index. This means we can’t add an element at the end of that sequence (which means we can’t add it to the place with the highest index – because there is no such place) which does not mean we can’t add it AFTER all of its elements (thereby creating a place with the highest index.)
You’re probably aware of the fact that I believe that to be the best definition of the word “infinite” out there – better than “without an end”. The reason being very simple: sets have no notion of “end” and a sequence can be infinite even if it has a beginning and an end.
The set of even integers is NOT greater than the set of integers. Nonetheless, the number of even integers is CERTAINLY greater than every integer. Are you saying that the number of even integers is an integer? If so, which one?
“Progressing pattern” is a figurative description of what a sequence is. It’s not a strict mathematical definition. And since sequences do not exist in time, they are neither increasing nor decreasing.
You choose where you’re going to insert it. You can insert it between two existing places e.g. you can insert it between the first and second element (thereby becoming second element.) But you can also insert it between one non-existent and one existent place e.g. you can add it between the zero-eth place (which doesn’t exist) and first place (thereby becoming first element.) Finally, you can insert it after all other elements (thereby becoming the last element.)
When adding (a) to ((1, 2, 3, \dotso)), I added it after all of its elements. The result is ((1, 2, 3, \dotso, a)).
And note that I added it to THE SAME sequence. I didn’t start a new one. That’s NOT the operation I performed.
You may want to argue such an operation is an impossible one rather than saying I added it somewhere I didn’t.