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修正 lint 规则名称和注释,确保一致性和准确性
Signed-off-by: szdytom <szdytom@qq.com>
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@ -5,13 +5,13 @@ enforce = true
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comment = "Found trailing whitespace at line end"
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[[rule]]
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name = "not-centered-dots-within-big-notations"
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pattern = "[=+][\\s&]*dots[^\\.]"
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name = "not-centered-dots-within-operators"
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pattern = "[=+(inter)(union)][\\s&]*dots[^\\.]"
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enforce = true
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comment = "Please use dots.c within = or +"
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comment = "Please use dots.c within operators"
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[[rule]]
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name = "centered-dots-within-small-notations"
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name = "centered-dots-within-commas"
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pattern = "[,][\\s&]*dots\\.c"
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enforce = true
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comment = "Please use dots within ,"
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@ -328,13 +328,13 @@
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#exercise_sol(type: "proof")[
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证明:$V$ 的任意一族子空间的交集是 $V$ 的子空间。
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][
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设 $V_1,dots,V_n$ 都是 $V$ 的子空间,记 $S=V_1 inter dots inter V_n$。我们关于 $n$ 使用数学归纳法。
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设 $V_1, dots, V_n$ 都是 $V$ 的子空间,记 $S=V_1 inter dots.c inter V_n$。我们关于 $n$ 使用数学归纳法。
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/ 第 $1$ 步: \
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当 $n=1$ 时,$S=V_1$,显然是 $V$ 的子空间。
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/ 第 $k$ 步: \
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假设当 $n = k - 1$ 时,结论成立,即 $V_1 inter dots inter V_(k - 1)$ 是 $V$ 的子空间。又因为 $V_k$ 是 $V$ 的子空间,由@E-inter-of-subspace-is-subspace 可知,$(V_1 inter dots inter V_(k - 1)) inter V_k$ 也是 $V$ 的子空间。由此,我们证明了当 $n = k$ 时,结论也成立。
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假设当 $n = k - 1$ 时,结论成立,即 $V_1 inter dots.c inter V_(k - 1)$ 是 $V$ 的子空间。又因为 $V_k$ 是 $V$ 的子空间,由@E-inter-of-subspace-is-subspace 可知,$(V_1 inter dots.c inter V_(k - 1)) inter V_k$ 也是 $V$ 的子空间。由此,我们证明了当 $n = k$ 时,结论也成立。
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#tab 综上所述,$V$ 的任意一族子空间的交集是 $V$ 的子空间。
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]
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