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I'm having a little trouble understanding how the limes inferior/superior are defined in Ahlfor's Complex Analysis.

Suppose we have a real sequence $\{\alpha_{n}\}_{1}^{\infty}$ and $a_{n} = \textrm{max} \{ \alpha_{1}, \alpha_{2}, \dots, \alpha_{n} \}$. The sequence $\{a_{n}\}_{1}^{\infty}$ is nondecreasing, hence it has a limit $\mu_{1}$ which is finite or $+\infty$.

Let $\mu_{k}$ be the least upper bound of the sequence $\{\alpha_{n}\}_{k}^{\infty}$ obtained from $\{\alpha_{n}\}_{1}^{\infty}$ by deleting $\alpha_{1},\dots,\alpha_{k-1}$.

Ahlfor's then states that $\{\mu_{k}\}$ is a nonincreasing sequence and its limit is given by $\mu$. I'm not quite convinced of this.

I've convinced myself that $\{\alpha_{n}\}_{1}^{\infty}$ is a nondecreasing sequence. If $\{\alpha_{n}\}_{1}^{\infty}$ is a nondecreasing sequence, then $\alpha_{1} \leq \alpha_{2} \leq \cdots \leq \alpha_{n}$, and this sequence has a limit, $\mu_{1}$. So when I construct the sequence $\{ \mu_{k} \} = \{ \mu_{1}, \mu_{2}, \dots\}$, it should be a nonincreasing sequence, which means $\mu_{1} \leq \mu_{2} \leq \cdots$

Thus, if $\{\alpha_{n}\}_{1}^{\infty} = \{ \alpha_{1},\alpha_{2},\dots\} \rightarrow \mu_{1}$, and $\{\alpha_{n}\}_{2}^{\infty} = \{ \alpha_{2},\alpha_{3},\dots\} \rightarrow \mu_{2}$, and so on and so forth, how is it that $\mu_{1} \leq \mu_{2} \leq \cdots$?

I'd appreciate any light you can bring to this issue. Thank you!

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    The sequence $\alpha_n$ doesn't have to be non-decreasing, $a_n$ has to be the one – 2017-01-18
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    Given any real sequece $x_n$ its either bounded above or not. If it is bounded above, then $\sup x_n$ is finite, otherwise you will find a subsequence $x_{n_k}\to \infty$ in which case $\sup x_n$ may be said to equal $+\infty$. Here $\mu_k$ is just $\sup_{n\ge k} \alpha_n$ which may equal $\infty$, and of course $\mu_k$ is then non-increasing – 2017-01-18
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    Whenever $A\subset B$ then $\sup(A)\leq \sup(B)$ and $\inf(A)\geq \inf(B)$, simply by element-wise inclusion. In your case $A=\left\{a_{k+1},a_{k+2},...\right\}$ and $B=\left\{a_{k},a_{k+1},a_{k+2},...\right\}$ – 2017-01-18
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    Thank you all! @user160738 I...actually hadn't thought of it that way, thank you! – 2017-01-18

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