MHB Converging Subsequences: Finding a Sequence for All Integers

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The discussion focuses on finding a sequence with subsequences that converge to every integer, including both positive and negative integers. A proposed sequence is 0, -1, 0, 1, -2, -1, 0, 1, 2, -3, -2, -1, 0, 1, 2, 3, which includes every integer infinitely often. This allows for the selection of constant subsequences that converge to any specified integer. The solution effectively addresses the challenge of including negative integers alongside positive ones. The proposed sequence successfully meets the original goal of convergence for all integers.
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I'm trying to find a sequence that has subsequences that converge to every integer. The question before that was the same but just for the positive integers, for which i gave {1,1,2,1,2,3...} but I'm struggling to include the negatives. Thanks
 
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Carla1985 said:
I'm trying to find a sequence that has subsequences that converge to every integer. The question before that was the same but just for the positive integers, for which i gave {1,1,2,1,2,3...} but I'm struggling to include the negatives. Thanks

Choose for example $$0,-1,0,1,-2,-1,0,1,2,-3,-2,-1,0,1,2,3,\ldots$$ and so on. Notice that every integer appears infinitely many times: we can choose a subsequence that converges to a given integer. Besides, that subsequence is constant.
 
That's fab, thank you! :)
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

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