SOS Henie归并性定理证明

╰☆ヾo.海x

这个有人称为海涅归并性定理。。 这只是其中一条，怎么证啊？ 我只知道跟连续性搭噶的那一条的证明。。

icesheep

右推左，反证法；
左推右，都用 epsilon-delta 语言写成定义就显然了。

hbghlyj

Heine definition of limit of a function at infinity using sequences
Suppose $\lim_{x\to+\infty}f(x)=L$. That is, for every $\epsilon>0$ there exists $x_0$ such that $x>x_0$ implies $|f(x)-L|<\epsilon$. Let $x_n\to+\infty$, that is for every $M>0$ exists $n_0\in\Bbb N$ such that $n>n_0$ implies $x_n>M$. I claim that $f(x_n)\to L$. Indeed, let $\epsilon>0$ be given. Then pick $x_0$ such that $x>x_0$ implies $|f(x)-L|<\epsilon$. Then pick $n_0\in\Bbb N$ such that $n>n_0$ implies $x>x_0$. It follows that $|f(x_n)-L|<\epsilon$ for $n>n_0$.

Next suppose $\lim_{n\to\infty}f(x_n)=L$ for all sequences with $x_n\to\infty$.
I claim that $\lim_{x\to+\infty}f(x)=L$.
Indeed, let $\epsilon>0$ be given.
Assume there does not exist $M$ such that $x>M$ implies $|f(x)-L|<\epsilon$. Then we can define a sequence $x_n$ as follows: For given $n\in \Bbb N$ pick $x_n$ arbitrary with $x_n>n$ and $|f(x_n)-L|\ge\epsilon$ (such $x_n$ exists by our assumption applied to $M=n$). Then clearly $x_n\to +\infty$ and hence $|f(x_n)-L|<\epsilon$ for sufficiently large $n$. As this contradicts $|f(x_n)-L|\ge\epsilon$, the initial assumption must be wrong. That is: There does exist some $M$ such that for all $x>M$ we have $|f(x)-L|<\epsilon$.

hbghlyj

Let $ℕ_∞$ be the one-point compactification of the discrete space $ℕ$ of natural numbers. Then a sequence $x_i$ in a space $X$ has a limit $c$ if and only if the (trivially continuous) map $x:ℕ → X$ extends to a continuous map $x̂:ℕ_∞ → X$ with $x̂(∞)=c$.
$$\xymatrix@C=2cm{
\mathbb N \ar@{^{(}->}[d]
&X\ar@{<-}[l]_{(x_i)}\ar@{<-}[ld]^{(x_i;c)}\\
\mathbb N_\infty}$$
Alexandroff extension

hbghlyj

→ is continuous, is equivalent to $\lim_{i\to\infty} x_i=c$
↓ is continuous function $f$
→ and ↓ are continuous, implies that ↘ is continuous, which is equivalent to $\lim_{i\to\infty} f(x_i)=\ell$
$$\xymatrix@C=2cm{
\mathbb N_\infty
&X\ar@{<-}[l]_{(x_i;c)}\ar@{->}[d]^{f}\\&
Y\ar@{<--}[lu]^{(f(x_i);f(c))}}$$