高考第一题, 集合论.

Czhang271828

(找个机会试一下 $\texttt{\\kongji}$ 这个command.)

下面四句话都是显然的
A $\kongji$ 到 $\{\kongji\}$ 有 1 个映射,
B $\kongji$ 到 $\kongji$ 有 1 个映射,
C $\{\kongji\}$ 到 $\kongji$ 有 0 个映射,
D $\{\kongji\}$ 到 $\{\kongji\}$ 有 1 个映射.

$|Y^X|=|Y|^{|X|}$ 有助于理解上述论断, 其中 $0^0=0$ 是自然的.

hbghlyj

en.wikipedia.org/wiki/Fibonacci_sequence#Combinatorial_proofs
$ F_{n} $ can be interpreted as the number of (possibly empty) sequences of 1s and 2s whose sum is $ n-1 $. This can be taken as the definition of $ F_{n} $ with the conventions $ F_{0}=0 $, meaning no such sequence exists whose sum is −1, and $ F_{1}=1 $, meaning the empty sequence "adds up" to 0. In the following, $ |{...}| $ is the cardinality of a set:
    $ F_{0}=0=|\{\}| $
    $ F_{1}=1=|\{\{\}\}| $
    $ F_{2}=1=|\{\{1\}\}| $
    $ F_{3}=2=|\{\{1,1\},\{2\}\}| $
    $ F_{4}=3=|\{\{1,1,1\},\{1,2\},\{2,1\}\}| $
    $ F_{5}=5=|\{\{1,1,1,1\},\{1,1,2\},\{1,2,1\},\{2,1,1\},\{2,2\}\}| $

hbghlyj

en.wikipedia.org/wiki/Zero_to_the_power_of_zero
Many widely used formulas involving natural-number exponents require $0^0$ to be defined as 1.  For example, the following three interpretations of $b^0$ make just as much sense for b = 0 as they do for positive integers $b$:

• The interpretation of $b^0$ as an empty product assigns it the value 1.
• The combinatorial interpretation of $b^0$ is the number of 0-tuples of elements from a b-element set; there is exactly one 0-tuple.
• The set-theoretic interpretation of $b^0$ is the number of functions from the empty set to a $b$-element set; there is exactly one such function, namely, the empty function.^[1]
  

All three of these specialize to give $0^0 = 1$.

[1] Bourbaki, Nicolas (2004). "III.§3.5". Elements of Mathematics, Theory of Sets. Springer-Verlag.

hbghlyj

Empty product
Empty sum
Why is the determinant of the 0x0 matrix equal 1? [closed]

hbghlyj

Czhang271828 发表于 2023-1-13 21:41
其中 $0^0=0$ 是自然的

Consider
$$ \lim_{t \to \infty} \left( e^{-t}\right)^\frac{1}{t} \to 0^0$$
But
$$\lim_{t \to \infty} \left( e^{-t}\right)^\frac{1}{t} = \lim_{t \to \infty} \left( e^{-t \frac{1}{t}}\right) = \lim_{t \to \infty} \left( e^{-1} \right) = \frac{1}{e} \neq 1$$
math.stackexchange.com/questions/1594063/