Let {X_n} seq.
點樣可以general地構造uncountable咁多個subseq?

Let P be power set of \mathbb{N}
Set of all subsequence of X_n = \{\{X_i\}_{i\inP}\}

自膠 應該係咁:
Let P be set of all infinite subset of \mathbb{N}
Set of all subsequence of X_n = \{\{X_i\}_{i\inP}\}

呢個方法work但係寫唔到出黎

留個名支持先

Let {X_n} seq.
點樣可以general地構造uncountable咁多個subseq?

Let P be power set of \mathbb{N}
Set of all subsequence of X_n = \{\{X_i\}_{i\inP}\}

自膠 應該係咁:
Let P be set of all infinite subset of \mathbb{N}
Set of all subsequence of X_n = \{\{X_i\}_{i\inP}\}

呢個方法work但係寫唔到出黎

[哲學題] 咩叫寫到出黎? [/哲學題]

無知問句
我係2013DSE 有修m2 本身對數都好有興趣 如果想了解下number theory 有冇書比新手 巴打介紹下
Sorry 1999

Let {X_n} seq.

點樣可以general地構造uncountable咁多個subseq?

我剩係諗到一個方法:
let P be set of prime
say S = Power set of P, then S is uncountable
[#FF0000]for {p_1,...,p_k} belongs to S
maps to seqs: {X_n} for n=p_i^j, 1<=i<=k, j is integer.[/#FF0000]
then the maps gives uncountable subseq.

仲有冇其他方法

當你一話個index set 係 `NN`時, 已經做緊一個 countable subsequence, 因為countable 或 uncountable 與否是看index set.

我係講緊uncountable咁多個subseq，唔係uncountable seq呀。
power set of integer唔work, 我講少左野, 要inf. subseq, not finite subseq.

我終於明白你講咩.... 根本係trivial.

你要明白咩係 subsequence.

Suppose there exists a function $\mathcal X: \mathbb N \to X$. 非常明顯, $\mathcal X = \left\{x_n \right\}_{n \in \mathbb N}$ is a sequence. 若然 $ \Psi: \mathbb N \to \mathbb N$ 係一個strictly increasing function, 根據定義 $\mathcal X \circ \Psi: \mathbb N \to X$ 都是一個sequence. 而我們就定義 $\mathcal X \circ \Psi$ 為 subsequence.

同時, 我們已知道$\#\psi$ 係 uncountable, where $\psi = \left\{ \Psi_i \forall i \right\}. 所以任何一個sequence 都有uncountably many subsequences.

Let {X_n} seq.

點樣可以general地構造uncountable咁多個subseq?

我剩係諗到一個方法:
let P be set of prime
say S = Power set of P, then S is uncountable
[#FF0000]for {p_1,...,p_k} belongs to S
maps to seqs: {X_n} for n=p_i^j, 1<=i<=k, j is integer.[/#FF0000]
then the maps gives uncountable subseq.

仲有冇其他方法

當你一話個index set 係 `NN`時, 已經做緊一個 countable subsequence, 因為countable 或 uncountable 與否是看index set.

我係講緊uncountable咁多個subseq，唔係uncountable seq呀。
power set of integer唔work, 我講少左野, 要inf. subseq, not finite subseq.

我終於明白你講咩.... 根本係trivial.

你要明白咩係 subsequence.

Suppose there exists a function $\mathcal X: \mathbb N \to X$. 非常明顯, $\mathcal X = \left\{x_n \right\}_{n \in \mathbb N}$ is a sequence. 若然 $ \Psi: \mathbb N \to \mathbb N$ 係一個strictly increasing function, 根據定義 $\mathcal X \circ \Psi: \mathbb N \to X$ 都是一個sequence. 而我們就定義 $\mathcal X \circ \Psi$ 為 subsequence.

同時, 我們已知道$\#\psi$ 係 uncountable, where $\psi = \left\{ \Psi_i \forall i \right\}$. 所以任何一個sequence 都有uncountably many subsequences.

無知問句
我係2013DSE 有修m2 本身對數都好有興趣 如果想了解下number theory 有冇書比新手 巴打介紹下
Sorry 1999

教科書:
Niven, Zuckerman, Montgomery, An Introduction to the theory of numbers, 5th edition, John Wiley and Sons, Inc

呢本有D難:
The Queen of Mathematics: A Historically Motivated Guide to Number Theory

無知問句
我係2013DSE 有修m2 本身對數都好有興趣 如果想了解下number theory 有冇書比新手 巴打介紹下
Sorry 1999

有explanations, 有 solutions, 有大量examples, 適合任何人士閱讀。

其實, 我知你想講 $\bar \mathbb R$, two-point compactification of `RR`. 但這個只是為方便, 例 我話一個sequence diverge, 但又可以話一個sequence converges in `bar RR`，根本無實際意義。另外，如measure theory, Beppo Levi theorem 就是要用 `bar RR`, 免得要分多幾個cases。 其實, 你可以話係懶, 多過有特別意義。

好多 function 會 d 左個 infinity 出黎
例如 sqrt(x) at x = 0 既 derivative 係 infinity
所以D人研究 Lp spaces 避唔開要放埋 infinity 落去

你可否講清楚你是講緊 $L^p$ Lebesgue space 還是 $W^{1, p}$ Sobolev space 呢?

你可否講清楚你是講緊 $L^p$ Lebesgue space 還是 $W^{1, p}$ Sobolev space 呢?

$L^p$ space. Anything wrong?