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On 11.12.2024 21:58, Jim Burns wrote:On 12/11/2024 2:57 PM, WM wrote:On 11.12.2024 20:27, Jim Burns wrote:
Each "leaves" by
not.being.in.common.with.all.end.segments.
∀k ∈ ℕ:
k ∉ E(k)\{k} = E(k+1) ⊇ ⋂{E(i):i} ∌ k
⋂{E(i):i} = {}.
>Therefore,>
one.element.emptier ℕ\{0}
is not.smaller.than ℕ
It is a smaller set.
For each k in ℕ
there is unique k+1 in ℕ\{0}
>Cardinalities are not useful.>
And yet, by ignoring them,
you (WM) end up wrong about
⎛ For each k in ℕ
⎝ there is unique k+1 in ℕ\{0}
Is the complete removal of natural numbers from the sequence of intersectionsEach "leaves" by
bound by the law
∀k ∈ ℕ :
∩{E(1), E(2), ..., E(k+1)} =
∩{E(1), E(2), ..., E(k)} \ {k}
or not?
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