Problem 1727

1. Assume the opposite: Suppose that the interval $I$ can be decomposed into finitely many mutually disjoint proper subsets which are all congruent by translation. That is, $I = \bigcup_{i=1}^m A_i$, where $A_i \cap A_j = \varnothing$ for $i \neq j$ and $A_i = A_1 + t_i$ for $i > 1$. Since the subsets are proper, $m > 1$. Without loss of generality (WLOG), assume that the left endpoint of $I$ is $0$ and $1 = t_2 < t_3 < \dots < t_m$.

2. Closed Interval Case: Let $I$ be a closed interval. For any $n \in \mathbb{N}$, denote $P_n = \{i \mid A_i \cap [0, n) \neq \varnothing\}$. For example, $P_1 = \{1\}$ because $A_1$ is the "leftmost" of all sets and $0$ may belong only to it, like any other point before the closest "translation", which is $t_2 = 1$ belonging to $A_2$. It is also clear that $[1, 2) \subseteq A_2$ as a translation of $[0, 1) \subseteq A_1$. Thus, $P_2 = \{1, 2\}$.

3. Induction Hypothesis: We prove by induction the following statements simultaneously:
- (1) For any $n \in \mathbb{N}$, $[0, n) \subseteq I$.
- (2) Given any $i = 1, 2, \dots, n$, all points in $[i-1, i)$ belong to the same set.
- (3) If $i \in P_n$ for $i > 1$, then $t_i \in \mathbb{N}$.

4. Base Case: The induction base and the case $n = 2$ have already been proven.

5. Induction Step: Let the statements be true for some $n \geq 2$. Since $I$ is a closed set, (1) implies $n \in I$. Let $n \in A_j$ and consider three possible cases:
- Case 1: $j \notin P_n$. Then $t_j = n$ and $[n, n+1) \subseteq A_j$ as a translation of $[0, 1) \subseteq A_1$.
- Case 2: $j \in P_n$ and $j > 1$. Then $t_j \in \mathbb{N}$, as follows from (3) and, like in the previous case, $[n, n+1) \subseteq A_j$ as a translation of $[n-t_j, n+1-t_j) \subseteq A_1$.
- Case 3: $j = 1$. Then $n+1 \in A_2$. To prove that $[n, n+1) \subset A_1$, take any point $x$ from $(n, n+1)$ and let $x \in A_k$. The case $k \notin P_n$ is impossible because $A_k$ would contain some interval of length $1$ (namely $[t_k, t_k+1)$) embracing $n+1$ along with $x$. The case $k \in P_n, k \ne 1$ is also impossible because in this case the whole $[n, n+1)$ would belong to $A_k$ (see Case 2), which contradicts $n \in A_1$. Thus, $k = 1$.

6. Conclusion for Closed Interval: We've just proved the three statements. From the first statement, it follows that $I = [0, +\infty)$, which is a contradiction.

7. Open Interval Case: Let $I$ be an open interval. For any $n \in \mathbb{N}$, denote $P_n = \{i \mid A_i \cap (0, n] \neq \varnothing\}$. Note that no point $x \in (0, 1]$ may belong to $A_i$ other than $A_1$ because in that case $y = x - t_i$ must belong to $A_1 \subseteq I$, but $y \leq 0$ so $y \notin I$. On the other hand, at least some right neighborhood $(0, \varepsilon) \subseteq I$, so there is a point $x \in (0, 1)$ belonging to $A_1$. This implies $x + 1 \in A_2 \implies (0, 1] \subset I \implies (0, 1] \subseteq A_1$. Thus, $P_1 = \{1\}$. It is also clear that $(1, 2] = (0 + t_1, 1 + t_1] \subseteq A_2$ and $P_2 = \{1, 2\}$.

8. Induction Hypothesis for Open Interval: We prove by induction the following statements simultaneously:
- (1) For any $n \in \mathbb{N}$, $(0, n] \subseteq I$.
- (2) Given any $i = 1, 2, \dots, n$, all points in $(i-1, i]$ belong to the same set.
- (3) If $i \in P_n$ for $i > 1$, then $t_i \in \mathbb{N}$.

9. Base Case for Open Interval: The induction base and the case $n = 2$ have already been proven.

10. Induction Step for Open Interval: Let the statements be true for some $n \geq 2$. Since $I$ is an open set, (1) implies that at least some right neighborhood $(n, n + \varepsilon) \subseteq I$. Consider three possible cases:
- Case 1: There is $x \in (n, n+1]$ belonging to some $A_j$ such that $j \in P_n$ and $j > 1$. Then $x - t_j \in A_1$. (3) says that $t_j \in \mathbb{N}$, while (2) implies $(n - t_j, n + 1 - t_j] \subseteq A_1$, thus $(n, n + 1] \subseteq A_j$.
- Case 2: There is $\varepsilon > 0$ such that $(n, n + \varepsilon) \subset A_1$. Then $(n + 1, n + 1 + \varepsilon) \subset A_2$. It entails two things: (a) $(n, n + 1] \subset I$; (b) none of $x \in (n, n + 1]$ belong to $A_j$ with $j \notin P_n$ because the latter would mean $n + \varepsilon \leq t_j < x \leq n + 1$ and $A_j$ contained some interval of length $1$ (namely $(t_j, t_j + 1]$) embracing $(n + 1, n + 1 + \varepsilon)$ along with $x$. From (a) and (b) it follows that $(n, n + 1] \subset A_1$.
- Case 3: For any $\varepsilon > 0$, there is $x \in (n, n + \varepsilon)$ such that $x \notin A_1$. It means that there is $j \notin P_n$ such that for any $\varepsilon > 0$, $(n, n + \varepsilon) \cap A_j \neq \varnothing$. But then $t_j = n$ and $(n, n + 1] \subseteq A_j$.

11. Conclusion for Open Interval: We've just proved the three statements. From the first statement, it follows that $I = (0, +\infty)$, which is a contradiction.