What is meant by the statement and the proof of Lemma 3.2 in Chapter XII of Dehornoy's book Braids and Self-Distributivity?

That lemma states "Assume that $j_1$ and $j_2$ are elementary embeddings of ($R_\lambda$ , $\in$) into itself. Then so is $j_1[j_2]$."

Here, $j_1[j_2]$ was defined in Definition 3.1, where $j_1$ and $j_2$ were elementary embeddings of the limit rank $R_\lambda$ into itself (by "$R_\lambda$" Dehornoy means what other authors call $V_\lambda$, $\lambda$ an ordinal).

$\bigcup_{\gamma\lt\lambda} j_1(j_2\big| R_\gamma)$

The difficulty I have here is that definition 3.1 does not actually define an elementary embedding of $R_\lambda$ into itself. $j_2\big| R_\gamma$ is a set of ordered pairs whose first element ranges over $R_\gamma$, and so $j_1$ of that is a set of ordered pairs whose first element only occurs in the image of $j_1$. In particular, the elementary embedding being defined is not defined for elements of $R_\lambda$ not in the range of $j_1$.

I may have seen some other authors suggest that one can define $j_1[j_2]$ to be the identity on the complement of the range of $j_1$, but Dehornoy does not do this, and this convention would not be consistent with the terminology in the proof of Lemma 3.2 besides.

Similarly, the proof of lemma 3.2 just does not make sense for the same reason.

Edit: Here are some notes based on the answer and comments. Perhaps they will be helpful to future readers.

Let $j$ be an elementary embedding from $R_\lambda$ to $R_\lambda$. Let $X$ be a proper subset of $R_\lambda$. Let $\phi$ be the formula $x\in y$. For each $a$ in $X$, $\phi(a, X)$ holds, and so $j(a)$ is in $j(X)$. Let $Y=\lbrace j(x):x\in X\rbrace$, the pointwise application of $j$ to $X$. Then $j(X)=Y\cup K$ where $K$ does not intersect the range of $j$. If $X$ is finite then $K$ is empty. If $X\subset X'$ then $K\subset K'$ where $K'$ corresponds to $X'$ as $K$ does to $X$.

If $X$ is an ordered pair $(x,y)$ then $j(x)$ is the ordered pair $(j(x), j(y))$.

Hence, if $X$ is a function, then $j(X)$ is the function $Y\cup K$, where $Y$ is the pointwise application of $j$ to $X$ consisting of all ordered pairs $(j(x),j(y)):(x,y)\in X)$. $K$ is also a function whose domain does not intersect the range of $j$.

Now, $j_2$ is the set of ordered pairs $(x,j_2(x))$ for $x$ in $R_\lambda$. Hence, $j_1[j_2]$ is the union of the set $\lbrace (j_1(x), j_1(j_2(x)): x\in R_\lambda$ with a function $K$ whose domain does not intersect the range of $j_1$.

This shows at once that for $z$ in the range of $j_1$, that $j_1[j_2](z)=j_1(j_2(j_1^{-1}(z)))$, motivating the self-distributivity of $[]$.

As to the purpose and structure of these $K$, I have no insight.

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    $\begingroup$ If $f$ is a function from $V_\alpha$ to $V_\beta$, then $j_1(f)$ is a function from $V_{j_1(\alpha)}$ to $V_{j_1(\beta)}$ (not from the image $j_1[V_\alpha]$). $\endgroup$ Commented Dec 22, 2021 at 9:53

1 Answer 1


As Monroe Eskew points out in comments: $j_1(j_2|_{V_\gamma})$ really means $j_1$ applied to $j_2|_{V_\gamma}$ (not the pointwise image of $j_2|_{V_\gamma}$ under $j_1$, as you seem to be reading it). So since $j_2|_{V_\gamma}$ is a function $V_\gamma \to V_{j_2(\gamma)}$, you get that $j_1(j_2|_{V_\gamma})$ is a function $V_{j_1(\gamma)} \to V_{j_1(j_2(\gamma))}$, as required.

  • $\begingroup$ Thank you for the clearing up a number of misconceptions and thanks @monroe-eskew. Note that because $j_1(\gamma)\ge\gamma)$, $R_{j_1(\gamma)}$ contains $R_\gamma$ which insures $j_1$ is total on $R_\lambda$. These elementary embeddings of set theory or huge parts of it really take some getting used to. $\endgroup$
    – kdog
    Commented Dec 23, 2021 at 19:47
  • $\begingroup$ I added some notes to my question based on this answer. I think this explains also why I confused $j_1(j_2\mid R_\gamma)$ with the pointwise image. $\endgroup$
    – kdog
    Commented Dec 24, 2021 at 6:43

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