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Let $I$ be an uncountable well-ordered set and let $A$ be the set of all $a \in I$ such that the set of all elements smaller than $a$ is uncountable. Do we always have $A\neq \emptyset$ and how does one prove this?

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There is exactly one counterexample (up to order isomorphism), namely taking $I$ to be $\omega_1$.

If, without loss of generality, we take $I$ to be an uncountable ordinal $\alpha$, then your $A$ is the interval $[\omega_1,\alpha)$, which is empty exactly when $\alpha$ is $\omega_1$ itself.

(In more concrete terms, the order type of $\omega_1$ can be realized as the set of isomorphism classes of well-orderings of $\mathbb N$, ordered by the "is a prefix of" relation.)

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Without knowing much about ordinals, here is a way to solve this.

Fix any uncountable well-ordered set $(A,<)$.

Now consider the set $B=\{a\in A\mid\{b\in A\mid b<a\}\text{ is uncountable}\}$. If $B$ is empty, then we have a counterexample, if not then there is some $a$ which is a minimal element of $B$, therefore $\{b\in A\mid b<a\}$ is a well-ordered set which is uncountable, but there is no element of $B$ which has uncountably many elements below it.

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