The Moore colouring forbids cross-injections

Suppose $f:X_0\to Y$ is continuous and injective for an uncountable $X_0\subseteq X$. Delete the countable sets $X_0\cap (X\cap Y)$ and $f^{-1}(X\cap Y)$. After this deletion, every remaining $\alpha$ and $f(\alpha )$ are distinct and lie on opposite sides of $X\setminus Y$ and $Y\setminus X$. One of the two orientations $\alpha <f(\alpha )$ or $f(\alpha )<\alpha$ holds on an uncountable subfamily; retain it.

For each retained $\alpha$, continuity at $\alpha$ and the neighborhood $W_{f(\alpha )}\cap Y$ of $f(\alpha )$ give a basic clopen $U_{\alpha }\ni \alpha$ with $U_{\alpha }\subseteq f^{-1}(W_{f(\alpha )}\cap Y)$. Encode $U_{\alpha }$ by a finite support $F_{\alpha }\subseteq X$ and its membership-bit function. Add $\alpha$ to the support if necessary.

Apply F3 and then finite/countable pigeonhole thinning so that the $F_{\alpha }$ form a $\mathrm{\Delta }$-system with root $F$, all petals have one size $k>0$, the membership bits on the fixed root $F$ have one fixed vector, the membership bits on the increasingly enumerated petals have one fixed vector ${\chi }_0$, the root lies below every retained $\alpha$, and the order type of each petal together with $f(\alpha )$ is constant. These are separate finite thinnings: agreement of the petal pattern alone would not control the root coordinates. Because $X\cap Y$ is countable and the petals are disjoint, discard the countably many petals meeting $X\cap Y$. The families $\mathcal{A}=\{(F_{\alpha }\setminus F)\cup \{f(\alpha )\}:\alpha \}$ and $\mathcal{B}=\{\{\beta ,f(\beta )\}:\beta \}$ are therefore uncountable, fixed-size, and pairwise disjoint.

Enumerate each member of $\mathcal{A}$ and $\mathcal{B}$ increasingly. The uniform order types give an insertion coordinate $r\le k$ for $f(\alpha )$ in the first enumeration, a column $s<2$ occupied by $\beta$ in the second, and the other column $t<2$ occupied by $f(\beta )$. Define $\pi :k+1\to 2$ by $\pi (r)=t$ and $\pi (i)=s$ for $i\mathrm{\ne }r$. Define the desired bit at row $r$ to be $0$, and at every other row to be the corresponding petal bit from ${\chi }_0$.

By [F2], choose $a\in \mathcal{A}$ and $b\in \mathcal{B}$ with $a<b$ that realize these bits. Let $a=(F_{\alpha }\setminus F)\cup \{f(\alpha )\}$ and $b=\{\beta ,f(\beta )\}$. At the inserted row, $c(f(\alpha ),f(\beta ))=0$, and $a<b$ ensures $f(\alpha )<f(\beta )$; hence $f(\beta )\mathrm{\notin }{W}_{f(\alpha )}$.

At every petal row, the realized bit says that $\beta$ satisfies the corresponding petal literal in the finite Boolean condition defining $U_{\alpha }$. At every root row, the separately stabilized root vector has the same value for $U_{\alpha }$ and $U_{\beta }$; since $\beta \in U_{\beta }$ and the root lies below $\beta$, F1 translates each root membership into precisely that fixed colouring bit. Thus every root and petal literal defining $U_{\alpha }$ holds at $\beta$, so $\beta \in U_{\alpha }$.

The containment chosen in step 2.1 now gives $f(\beta )\in W_{f(\alpha )}$, contradicting step 5.1. The construction used the orientation only to decide which column of the increasing pair is $\beta$; step 4.1 handles both orientations through $s,t$. Hence no such continuous injection exists.