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The question had a lot of views, and my answer didn't get any upvote. Now since I got a downvote, perhaps there was a (major) mistake somewhere in the argument I posted? Tbh I was reasonably confident about my answer

**[**since its not that unusual for accurate answers to go by with 0 upvotes

**]**, but the downvote def. confused me a bit.

I left a comment asking the reason for downvote, but it is not very likely that will get a response (would be good if it gets one). It is hard to tell whether the downvote is just impulse or based upon identification of an actual (major) mistake in answer. I would definitely like to know if its the latter.

I didn't think about the answer quite thoroughly (since I posted it after few hours of thinking), so maybe there is some aspect I missed? Nevertheless, I should say that the argument is of basic nature and it looked OK to me on few cursory/initial checks. But sometimes one can keep thinking thinking about along the same (wrong) lines when re-checking (especially when re-checks are not thorough enough). So a "second opinion"/"further input" would be quite helpful.

Link: https://mathoverflow.net/questions/381527/the-sum-of-two-well-ordered-subsets-is-well-ordered

Here is the question (not mine):

if ##A,B## are two well-ordered subsets of ##\mathbb{R}## (or any ordered group — with the induced order of course), the subset ##A+B:=\{a+b\,|\,a\in A,b\in B\}## is well-ordered. How does one see that?

Here is the answer I posted:

Let's suppose the well-order of ##A## and ##B## have order-types ##p## and ##q## respectively. So we use the notation ##a_i## (##i<p##) to denote the ##i##-th element of ##A##. We have ##a_i<a_j## whenever ##i<j<p##. Similarly, we use ##b_i## (##i<q##) to denote the ##i##-th element of ##B##.

Since ##A+B## is already linearly-ordered, we want to prove that ##A+B## has no infinite descent. In other words, we want to disprove the existence of a function ##f:\mathbb{N} \rightarrow \mathbb{R}## such that

**(i)**For all ##x## in domain of ##f## we have ##f(x)=a_i+b_j## (for some ##i<p## and ##j<q##)

**(ii)**##f## is a 1-1 function

**(iii)**For all ##i,j \in \mathbb{N}## (with ##j>i##) we must have ##f(j)<f(i)##.

So if we have ##f(x)=a_i+b_j## then we can write ##\mathrm{first}(f(x))## and ##\mathrm{second}(f(x))## to denote ##a_i## and ##b_j## respectively. Now we can define ##\alpha_0=min\{\alpha \in Ord:first(f(x))=a_{\alpha} \wedge x \in \mathbb{N} \}##. Let's denote ##n_0 \in \mathbb{N}## as the "last" value for which ##first(f(n_0))=a_{\alpha_0}##.

Now we define ##\alpha_1=min\{\alpha \in Ord:first(f(x))=a_{\alpha} \wedge x \in \mathbb{N} \wedge x>n_0 \}##. Let's denote ##n_1 \in \mathbb{N}## as the "last" value for which ##first(f(n_1))=a_{\alpha_1}##.

Now because of ##\alpha_1>\alpha_0##, we should get ##a_{\alpha_1}>a_{\alpha_0}## and hence ##second(f(n_1))<second(f(n_0))##. So it seems to me that when we define ##\alpha_2=min\{\alpha \in Ord:first(f(x))=a_{\alpha} \wedge x \in \mathbb{N} \wedge x>n_1 \}## and ##n_2## as the "last" value for which ##first(f(n_2))=a_{\alpha_2}##, then we should have similarly ##second(f(n_2))<second(f(n_1))##. The last inequality is supposed to follow from ##a_{\alpha_2}>a_{\alpha_1}## (because ##\alpha_2>\alpha_1##).

The previous paragraph seems to be suggestive of defining ##\alpha_i##, ##n_i## generally for all natural numbers ##i## and then creating an infinite descent for the second components:##......<second(f(n_3))<second(f(n_2))<second(f(n_1))<second(f(n_0))##.

And a comment regarding notation that I wrote:

It seems that instead of writing ##\alpha_1=min\{\alpha \in Ord:first(f(x))=a_{\alpha} \wedge x \in \mathbb{N} \wedge x>n_0 \}## it would probably be better to write something like: ##\alpha_1=min\{\alpha \in Ord: \exists x \in \mathbb{N}(first(f(x))=a_\alpha \wedge x>n_0)\}##