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$\newcommand{\N}{\mathbb{N}}$ Consider a bounded increasing infinite real sequence $(a_i)_{i \in \N}$. If I add an element $x$ larger than the supremum of $(a_i)$ to create another increasing sequence $(b_i)$, it contains the largest element $x$. Meanwhile, the index of every element is finite. Then, what can we say about the index of $x$? Is it also finite? I am confused because if it is finite, it will mean $(b_i)$ is finite. Please let me know what I am missing here. (I suspect I just can't make a new sequence in such a way.)

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    $\begingroup$ "Add it" where? To get an increasing sequence, you would need $k\geq i$ for all $i\in\mathbb{N}$. That's not a sequence any more, because a sequence is a function with domain $\mathbb{N}$, so every index has to be a natural number. So what you can say is "You can't add it to begin with." $\endgroup$ Oct 7, 2022 at 22:10
  • $\begingroup$ @ArturoMagidin You just excellently answered my question. Thank you. If you give an answer, I will accept it. $\endgroup$
    – Hermis14
    Oct 7, 2022 at 22:12
  • $\begingroup$ Relevant is Brian M. Scott's answer to: On defining sequences. $\endgroup$ Oct 7, 2022 at 22:17
  • $\begingroup$ "If I add an element $x$ larger than an upper bound of $(a_i)$ to have another increasing sequence $(b_i)$, $b_i$ has the largest element $x$." This sentence is incoherent. Please clarify. $\endgroup$ Oct 7, 2022 at 22:34
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    $\begingroup$ @AdamRubinson I thought Magidin has already explained why it is incoherent, didn't he? $\endgroup$
    – Hermis14
    Oct 7, 2022 at 22:37

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Recall that a real sequence is "really" a function $a\colon\mathbb{N}\to\mathbb{R}$; it's just that instead of writing $a(1)$, $a(2)$, etc. we write $a_1$, $a_2$, and so on; and because we can list the inputs, we can describe the whole function with a list, $(a_1,a_2,a_3,\ldots)$.

You start with an increasing bounded sequence, $(a_1,a_2,\ldots,a_n,\ldots)$, with $a_i\leq a_j$ whenever $i\leq j$, and with $a_k\leq M$ for all $k$. You say, "if I add an element $x$ greater than $M$" to the sequence, while keeping it an increasing sequence... but that is not really doable.

If you were to put this number $x$ in the sequence, it would have to be the image of some natural number $k$; and if the resulting sequence is still increasing, then because $a_m\leq x$ for all natural numbers $m$, you would need $m\leq k$ for all natural numbers $m$. There is no such natural number. So there is no way to "add" this term to the sequence: there is no place where you can place it (while keeping every term before its location where it is, and "shifting" every term after where you locate it one position over). The operation of adding a single term somewhere is doable. You just cannot do it and still keep an increasing sequence, if you are adding a term that is larger than all terms you already have.

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