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>> No.16189541 [View]
File: 245 KB, 1251x1433, 050346663cba134642ffd4b71465d3c2f.jpg [View same] [iqdb] [saucenao] [google]
16189541

>>16189533
>It's trivial. That's why the proof is so short.
I don't mean trivial as in its easy, I meant trivial as in it's not saying absolutely anything. Your definition is literally just "f has a limit at a if f is continuous at a".
>the pic you attached
Notice how [math]a \in \overline{A}[/math], the sequence is of points is in [math]A[/math] and the limit is with respect to [math]A[/math]. It's completely different from what you're talking about and obviously a generalization of the basic use case, which is [math]A = \mathbb{R} \setminus \{ a \}[/math].

>> No.15864774 [View]
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15864774

>>15864749
For any [math](a, b) \neq 0 \in \mathbb{Z}_p^2[/math], either [math]a = 0[/math], and then we can write [math](a, b) = b(0, 1)[/math] or [math]a \neq 0[/math] and we can write it as [math](a, b) = a (1, b/a)[/math].
This tells us that every vector is a multiple of one of [math]\{ (0, 1), (1, 0), (1, 1), \ldots, (1, p-1)[/math], so all the one-dimensional vector spaces are spans of one of those.

>> No.15126883 [View]
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15126883

What's the most realistic/achievable proposed design so far for a manned interstellar spaceship, anons?

>> No.11484600 [View]
File: 246 KB, 1251x1433, __remilia_scarlet_touhou_drawn_by_gotoh510__e828ac9528e1107febf6f55a5eefbfc2.jpg [View same] [iqdb] [saucenao] [google]
11484600

>>11484595
>that's neither true
>not true
>every vector has finite norm isn't true
You're working with some pants on head retarded definitions, aren't you?

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