云

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Examples of common false beliefs in mathematics
云(false belief)

云0.1. 注意到
\[\dim (U+V) = \dim U + \dim V - \dim U\cap V\]
我们有容斥原理\begin{align*}
\dim(U + V + W) &= \dim(U) + \dim(V) + \dim(W)\\
&\qquad\mathop{-} \dim(U \cap V) - \dim(U \cap W) - \dim(V \cap W)\\&\qquad \mathop{+} \dim(U \cap V \cap  W)
\end{align*}

这个公式是不成立的.(见the calculation of dim u v w)
只能做到\begin{align*} \dim(U +V + W) &= \dim((U +V) + W) \\ &= \dim(U +V) + \dim W - \dim((U+V)\cap W) \\ &=
\dim U + \dim V - \dim (U \cap V) + \dim W - \dim((U+V)\cap W) \end{align*}而无法继续套用二维情况的等式.

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云0.2. 由于
\[
\operatorname{Tr}(AB) = \operatorname{Tr}(BA),
\]
根据数学归纳法, 对$\tau \in S_k$,
\[
\operatorname{Tr}(A_1\cdots A_k) = \operatorname{Tr}(A_{\tau(1)}\cdots A_{\tau(k)}).
\]

云0.2的错误在于:由${\rm Tr}(AB)={\rm Tr}(BA)$,只能得到对$τ∈C_k,\operatorname{Tr}\left(A_{1} \cdots A_{k}\right)=\operatorname{Tr}\left(A_{\tau(1)} \cdots A_{\tau(k)}\right)$,而不能得到$τ∈S_k$时的结论.

维基百科:

Cyclic property
More generally, the trace is invariant under cyclic permutations, that is,
$$\operatorname{tr}(\mathbf{A}\mathbf{B}\mathbf{C}\mathbf{D}) = \operatorname{tr}(\mathbf{B}\mathbf{C}\mathbf{D}\mathbf{A}) = \operatorname{tr}(\mathbf{C}\mathbf{D}\mathbf{A}\mathbf{B}) = \operatorname{tr}(\mathbf{D}\mathbf{A}\mathbf{B}\mathbf{C}).$$
This is known as the ''cyclic property''.

Arbitrary permutations are not allowed: in general,
$$\operatorname{tr}(\mathbf{A}\mathbf{B}\mathbf{C}) \ne \operatorname{tr}(\mathbf{A}\mathbf{C}\mathbf{B}).$$

However, if products of three symmetric matrices are considered, any permutation is allowed, since:
$$\operatorname{tr}(\mathbf{A}\mathbf{B}\mathbf{C}) = \operatorname{tr}\left(\left(\mathbf{A}\mathbf{B}\mathbf{C}\right)^{\mathsf T}\right) = \operatorname{tr}(\mathbf{C}\mathbf{B}\mathbf{A}) = \operatorname{tr}(\mathbf{A}\mathbf{C}\mathbf{B}),$$
where the first equality is because the traces of a matrix and its transpose are equal. Note that this is not true in general for more than three factors.

hbghlyj

云 0.3. 设 F:C→D 为函子, 则 F 的像为 D 的子范畴.

证明. 由于 F(g∘f)=F(g)∘F(f), 我们仍有复合运算. 结合律与单位元显然.

云0.3的错误在于,
如下图,

F的像不是右图中的范畴的子范畴,因为F的像中有函子p,q却不包含它们的复合qp.(见MSE)

hbghlyj

云:
$\sin z$ 是有界函数，因为它是周期函数

这可以通过以下方式反驳：
(i) 有界的整函数是常数；
(ii) sinz 是整函数；
(iii) sinz 不是常数函数。

另一种方法是注意到 $\sin(iz)=i\sinh(z)$

hbghlyj

云:
实数轴的开且稠密子集 $U$ 必定是整个实数轴。
“证明”:
每个点 $x$ 都任意接近 $U$ 中的点 $u$，因此当在 $u$ 周围放置一个小邻域时，它必定包含 $x$.

这个推理的缺陷完全在于语言“任意接近”的模糊性，数学被语言模糊了。稠密集的定义：对于每个 x∈R 和每个 ε>0，存在某个 u∈U 使得 |x−u|<ε，但 x 本身不必在 U 中。

反驳：Non-trivial open dense subset of R

hbghlyj

threads.net/@haiiro.1023/post/DBIp5XrTN4h
女生讓男生厭惡的10個習慣:
(1)不溫柔
(2)認為有界閉集與緊集等價
(3)認為有限生成R模的子模也是有限生成的
(4)認為仿射簇的乘積的拓樸就是它們的Zariski拓樸的乘積拓樸
(5)認為流形上的聯絡都是無撓的
(6)算不清楚流形上的Laplace算子
(7)分不清投射模、內射模、平面模
(8)計算 Riemann曲率張量、Ricci張量時分不清上下指標,總是相差一個負號
(9)不熟悉自由生成、張量積、局部化等常見代數結構的泛性質。
(10)認為幾乎處處收斂一定是依測度收斂

hbghlyj

hbghlyj 发表于 2024-12-9 02:14
(3)認為有限生成R模的子模也是有限生成的

In general, submodules of finitely generated modules need not be finitely generated. As an example, consider the ring R = Z[X₁, X₂, ...] of all polynomials in countably many variables. R itself is a finitely generated R-module (with {1} as generating set). Consider the submodule K consisting of all those polynomials with zero constant term. Since every polynomial contains only finitely many terms whose coefficients are non-zero, the R-module K is not finitely generated.

Finitely generated module - Wikipedia