Let $f:[0,1]\rightarrow [0,1]$ be a continuous and bijective function.
Describe the set:
\[A=\{f(x)-f(y)\mid x,y\in[0,1]\backslash\mathbb{Q}\}\]
[hide="Note"]
You are given the result that [i]there is no one-to-one function between the irrational numbers and $\mathbb{Q}$.[/i][/hide]
1. **Claim**: The set \( A = \{ f(x) - f(y) \mid x, y \in [0,1] \setminus \mathbb{Q} \} \) is equal to \((-1, 1)\).

2. **Subset Inclusion**: First, we show that \( A \subseteq (-1, 1) \). Since \( f \) maps \([0,1]\) to \([0,1]\), for any \( x, y \in [0,1] \), we have \( 0 \leq f(x), f(y) \leq 1 \). Therefore, 
   \[
   -1 \leq f(x) - f(y) \leq 1.
   \]
   However, since \( f \) is continuous and bijective, it attains all values in \([0,1]\) exactly once. Thus, for \( x \neq y \), \( f(x) \neq f(y) \), and hence \( f(x) - f(y) \neq 0 \). This implies that \( f(x) - f(y) \) lies strictly within \((-1, 1)\).

3. **Non-Empty Set**: Clearly, \( 0 \in A \) because for \( x = y \), \( f(x) - f(y) = 0 \).

4. **Density Argument**: To show that every point in \((-1, 1)\) can be achieved, consider any \( r \in (0, 1) \). Let \( s = \min(r, 1 - r) \). We need to show that there exist \( x, y \in (0, s) \) such that \( f^{-1}(x), f^{-1}(x + r) \in \mathbb{R} \setminus \mathbb{Q} \).

5. **Contradiction**: Suppose for contradiction that for all \( x \in (0, s) \), either \( f^{-1}(x) \in \mathbb{Q} \) or \( f^{-1}(x + r) \in \mathbb{Q} \). This would imply the existence of an injection \( j: (0, s) \hookrightarrow \mathbb{Q} \) where \( j(x) \) returns the smallest element in \( \{ f^{-1}(x), f^{-1}(x + r) \} \cap \mathbb{Q} \).

6. **Cantor–Schröder–Bernstein Lemma**: Since \((0, s) \cong \mathbb{R}\) (there exists a bijection between \((0, s)\) and \(\mathbb{R}\)), we can find an injection \( k: \mathbb{R} \hookrightarrow (0, s) \). Combining these, we get an injection \( j \circ k: \mathbb{R} \hookrightarrow \mathbb{Q} \).

7. **Contradiction**: This contradicts the known result that there is no injection from \(\mathbb{R}\) to \(\mathbb{Q}\), as \(\mathbb{Q}\) is countable and \(\mathbb{R}\) is uncountable.

8. **Conclusion**: Therefore, our assumption must be false, and there must exist \( x, y \in (0, s) \) such that \( f^{-1}(x), f^{-1}(x + r) \in \mathbb{R} \setminus \mathbb{Q} \). This implies that \( f(x) - f(y) \) can take any value in \((-1, 1)\).

Thus, we have shown that \( A = (-1, 1) \).

The final answer is \( \boxed{ (-1, 1) } \).