## Return to Answer

4 the editings overlapped, apparently, it should be fixed now, sorry...

Here is a quick sketch (probably it can be made much cleaner). Using an inductive argument it should not be difficult to reduce to studying the case that `$(X,<_1)$` is isomorphic to a cardinal, say $\kappa$. For convenience, let us identify $(\kappa,<)$ and `$(X,<_1)$`. Let us now construct $Y$ using transfinite induction. Let $X'\subseteq X$ be the initial segment of $X$ (with respect to `$<_2$`) which, endowed with the order `$<_2$`, is isomorphic to $\kappa$. For `$\alpha<k$` define `$$y_\alpha=\min\{\beta\in X': \forall \alpha'<\alpha\; \beta> y_{\alpha'}\textrm{\ and\ }\beta>_2 y_{\alpha'}\}.$$` The set above is not empty, and so $y_\alpha$ is well defined, because `$|\{\gamma\in X':\exists \alpha'<\alpha\; \gamma\leq y_{\alpha}\}|y_{\alpha'}\}|<k$` and also `$|\{\gamma\in X':\exists \alpha'<\alpha\; \gamma\leq_2 y_\alpha\}|y_{\alpha'}\}|<k$` (this depends on `$\{y_{\alpha'}\}_\{alpha'\{y_{\alpha'}\}_{\alpha'<\alpha}$` not being cofinal in $(k,<)$ and `$(X',<_2)$`), so their complements in $X'$ intersect. The set `$Y=\{y_\alpha\}_{\alpha<k}$` is what we were looking for.

3 fixed some of the LaTeX by adding backticks, but gave up on one part

Here is a quick sketch (probably it can be made much cleaner). Using an inductive argument it should not be difficult to reduce to studying the case that `$(X,<_1)$` is isomorphic to a cardinal, say $\kappa$. For convenience, let us identify $(\kappa,<)$ and `$(X,<_1)$`. Let us now construct $Y$ using transfinite induction. Let $X'\subseteq X$ be the initial segment of $X$ (with respect to `$<_2$`) which, endowed with the order `$<_2$`, is isomorphic to $\kappa$. For `$\alpha<k$` define `$$y_\alpha=\min\{\beta\in X': \forall \alpha'<\alpha,\; <\alpha\; \beta> y_{\alpha'}\textrm{ y_{\alpha'}\textrm{\ and\ }\beta>_2 y_{\alpha'}\}.$$` The set above is not empty, and so $y_\alpha$ is well defined, because `$|\{\gamma\in X':\exists \alpha'<\alpha\; \gamma\leq y_{\alpha}\}|<k$` and also `$|\{\gamma\in X':\exists \alpha'<\alpha\; \gamma\leq_2 y_\alpha\}|<k$` (this depends on `$\{y_{\alpha'}\}_{\alpha'\{y_{\alpha'}\}_\{alpha'<\alpha}$` not being cofinal in $(k,<)$ and `$(X',<_2)$`), so their complements in $X'$ intersect. The set `$Y=\{y_\alpha\}_{\alpha<k}$` is what we were looking for.

2 fixed formatting problems

Here is a quick sketch (probably it can be made much cleaner). Using an inductive argument it should not be difficult to reduce to studying the case that `$(X,<_1)$` is isomorphic to a cardinal, say $\kappa$. For convenience, let us identify $(\kappa,<)$ and `$(X,<_1)$. <_1)$`. Let us now construct $Y$ using transfinite induction. Let $X'\subseteq X$ be the initial segment of $X$ (with respect to `$<_2$) <_2$`) which, endowed with the order `$<2$<_2$`, is isomorphic to $\kappa$. For `$\alpha\alpha=\min{\beta\in \alpha<k$` define ```$$y_\alpha=\min\{\beta\in X': \forall \alpha'<\alpha\; beta> y_{\alpha'}\textrm{<\alpha,\; \beta> y_{\alpha'}\textrm{ and \ }\beta>2 y{\alpha'}}.$$
\beta>_2 y_{\alpha'}\}.$$``` The set above is not empty, and so $y_\alpha$ is well defined, because `$|{\gamma\in |\{\gamma\in X':\exists \alpha'<\alpha\; \gamma\leq y_{\alpha}}|2)$)y_{\alpha}\}|<k$` and also `$|\{\gamma\in X':\exists \alpha'<\alpha\; \gamma\leq_2 y_\alpha\}|<k$` (this depends on `$\{y_{\alpha'}\}_{\alpha'<\alpha}$` not being cofinal in $(k,<)$ and `$(X',<_2)$`), so their complements in $X'$ intersect. The set `$Y={y````\alpha}_{\alpha
Y=\{y_\alpha\}_{\alpha<k}$``` is what we were looking for.

1

Here is a quick sketch (probably it can be made much cleaner). Using an inductive argument it should not be difficult to reduce to studying the case that $(X,<_1)$ is isomorphic to a cardinal, say $\kappa$. For convenience, let us identify $(\kappa,<)$ and $(X,<_1)$. Let us now construct $Y$ using transfinite induction. Let $X'\subseteq X$ be the initial segment of $X$ (with respect to $<_2$) which, endowed with the order $<2$, is isomorphic to $\kappa$. For $\alpha\alpha=\min{\beta\in X': \forall \alpha'<\alpha\; beta> y_{\alpha'}\textrm{\ and\ }\beta>2 y{\alpha'}}.$$The set above is not empty, and so$y_\alpha$is well defined, because$|{\gamma\in X':\exists \alpha'<\alpha\; \gamma\leq y_{\alpha}}|2)$), so their complements in$X'$intersect. The set$Y={y\alpha}_{\alpha
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