0
$\begingroup$

If $f:\kappa \to \kappa$ is strictly increasing and continuous, why then holds that

for every $\alpha < \kappa$, $f(\alpha) \geq \alpha$

Furthermore, why in this case also holds that $f[\kappa]$ is closed in $\kappa$? Where by closed I mean the following definiton:

A subset $C \subset \kappa$ is said to be closed in $\kappa$ if $\forall\lambda<\kappa(\lambda$ is a limit ordinal $\wedge$ $C\cap\lambda$ is unbounded in $\lambda \to \lambda \in C$)

  • 0
    I am trying to use the fact that $f$ is strictly increasing, but I don't see how the first inequality still holds.. – 2017-02-21
  • 1
    HINT: use induction. What can you say about the first $\alpha$ such that $f(\alpha)<\alpha$? – 2017-02-21
  • 0
    @NoahSchweber Can I do it like this: for $\alpha = 0,$ $f(0)\geq0$, obvious. For succesor case, $f(\alpha + 1)>f(\alpha)\geq\alpha$, so $f(\alpha+1)\geq\alpha+1$ and for limit case, $f(\alpha)=f(\cup_{\gamma<\alpha}\gamma)=\cup_{\gamma<\alpha}f(\gamma)\geq\cup_{\gamma<\alpha}\gamma=\alpha$ – 2017-02-21
  • 1
    Bingo! Although actually your limit step can be handled *without invoking continuity* - do you see how? (Continuity is only needed to show $f[\kappa]$ is closed.) – 2017-02-21
  • 0
    Yes I do and thanks for the hint. :) But I find hard for intuition this closedness-definition, which is why I still don't get how to prove the second part of my question.. – 2017-02-21
  • 1
    Have you seen my answer to your question about closedness? Did it make sense? – 2017-02-21
  • 0
    Yes it did, but where exactly do you use continuity? – 2017-02-21
  • 1
    Well, you want to show that $f[\kappa]$ is closed. So suppose $\lambda$ is such that $f[\kappa]\cap \lambda$ is unbounded in $\lambda$; we want to show $\lambda\in f[\kappa]$. Now, continuity lets us compute $f(\eta)$ (for $\eta$ limit) if we know $f(\beta)$ for "lots" (specifically, unboundedly many) $\beta<\eta$. So, do you see how to get a set of $\beta$s which will guarantee $f(\eta)=\lambda$ for some $\eta$? (HINT: think about the $\gamma$s in $f[\kappa]\cap \lambda$ . . .) – 2017-02-21
  • 0
    Taking for $\eta$ the supremum over all $\beta<\kappa$ such that $f(\beta)<\lambda$? – 2017-02-21
  • 1
    Bingo. Since we've worked everything out, I'll add an answer below so this question can be resolved. (Actually there's one subtlety left, see my answer.) – 2017-02-21

1 Answers 1

0

The following solution emerged in the comments:

  • To show that $f(\alpha)\ge\alpha$, we use the fact that $f$ is increasing (continuity is not needed here). Let, for contradiction, $\beta$ be the least ordinal such that $f(\beta)<\beta$. Then for each $\gamma<\beta$, we have $f(\gamma)\ge \gamma$; so since $f$ is increasing, we need to have $f(\alpha)>\gamma$ for all $\gamma<\alpha$. So $f(\alpha)\ge\alpha$.

  • To show that $f[\kappa]$ is closed, we use the continuity of $f$ (increasingness is not needed here). Let $\lambda<\kappa$ be such that $f[\kappa]\cap\lambda$ is unbounded in $\lambda$; then let $S=\{\eta: f(\eta)<\lambda\}$ be the preimages of the values of $f$ below $\kappa$. We now pick the least $\beta$ such that $S\cap \beta$ is unbounded in $\lambda$ (this trick was not mentioned in the comments - do you see why it's necessary? - but we can get around it if you want by using the fact that $f$ is increasing); it's not hard to show that continuity then forces $f(\beta)=\lambda$.

  • 0
    But if we take supremum as I first mentioned, and put $\eta:=sup${$\beta<\kappa | f(\beta)<\lambda$}=$\cup_{f(\beta)<\lambda}\beta$, don't we already get that $f(\eta)=\cup_{f(\beta)<\lambda}f(\beta)=\lambda$? – 2017-02-21
  • 1
    @Eurydice Not necessarily! For instance, consider the map $f:\omega+\omega+1\rightarrow ON$ given by: $f(n)=\omega+n$ for $n$ finite, $f(\omega)=\omega+\omega$, and $f(\alpha)=7$ for $\alpha>\omega$. This is continuous, even though $f(\omega+\omega)=7$ is less than you'd expect; do you see why? – 2017-02-21