For real positive numbers $x,$ the set $A(x)$ is defined by 
\[A(x)=\{[nx]\ | \ n\in{ \mathbb{N}}\},\] where $[r]$ denotes the greatest integer not exceeding real numbers $r.$Find all irrational numbers $\alpha >1$ satisfying the following condition.
Condition: If  positive real number $\beta$ satisfies $A(\alpha)\supset A(\beta),$ then $\frac{\beta}{\alpha}$ is integer.
1. **Define the problem and the set \( A(x) \):**
   For a real positive number \( x \), the set \( A(x) \) is defined as:
   \[
   A(x) = \{ [nx] \mid n \in \mathbb{N} \}
   \]
   where \( [r] \) denotes the greatest integer not exceeding the real number \( r \).

2. **Condition to be satisfied:**
   We need to find all irrational numbers \( \alpha > 1 \) such that if a positive real number \( \beta \) satisfies \( A(\alpha) \supset A(\beta) \), then \( \frac{\beta}{\alpha} \) is an integer.

3. **Using Beatty's Theorem:**
   According to Beatty's theorem, for irrational numbers \( \alpha \) and \( \beta \) such that \( \frac{1}{\alpha} + \frac{1}{\beta} = 1 \), the sequences \( \{ [n\alpha] \} \) and \( \{ [n\beta] \} \) partition the natural numbers \( \mathbb{N} \).

4. **Assume \( \alpha \) and \( \beta \) are such that \( A(\alpha) \supset A(\beta) \):**
   If \( A(\alpha) \supset A(\beta) \), then every element of \( A(\beta) \) must also be an element of \( A(\alpha) \). This implies that for every \( n \in \mathbb{N} \), there exists an integer \( k \) such that:
   \[
   [n\beta] = [k\alpha]
   \]

5. **Analyzing the condition \( \frac{\beta}{\alpha} \) being an integer:**
   Suppose \( \frac{\beta}{\alpha} = q \) where \( q \) is an integer. Then:
   \[
   \beta = q\alpha
   \]
   For any \( n \in \mathbb{N} \):
   \[
   [n\beta] = [nq\alpha]
   \]
   Since \( q \) is an integer, \( nq \) is also an integer, and thus:
   \[
   [nq\alpha] = nq[\alpha]
   \]
   This implies that \( A(\beta) \subset A(\alpha) \).

6. **Proving the necessity of \( \alpha > 2 \):**
   If \( \alpha \leq 2 \), then there exists a \( \beta \) such that \( \frac{\beta}{\alpha} \) is not an integer, contradicting the condition. Therefore, \( \alpha > 2 \).

7. **Conclusion:**
   All irrational numbers \( \alpha > 2 \) satisfy the condition that if \( A(\alpha) \supset A(\beta) \), then \( \frac{\beta}{\alpha} \) is an integer.

\(\blacksquare\)

The final answer is all irrational numbers \( \boxed{ \alpha > 2 } \).