i need help.. im not really sure how to start.
There exist an open interval A=(x,y), x set B, and a function f:A-->B such that f(A)≠ f(x),f(y).. prove
To prove that there exists an open interval A=(x,y), x ∈ B, and a function f:A --> B such that f(A) ≠ f(x), f(y), you need to follow these steps:
1. Start by assuming that there exists an open interval A=(x,y) and a function f:A --> B such that f(A) = f(x) = f(y).
2. Now, you need to show a contradiction to disprove this assumption. To do this, you can use the properties of open intervals and functions.
3. Recall that an open interval A=(x,y) includes all the numbers greater than x and less than y. So, we have a range of values between x and y.
4. Next, consider the function f:A --> B, which maps the values in A to the values in the set B. Since f(x) = f(y), it implies that the function has the same output for both x and y within the interval A.
5. To create a contradiction, choose a specific value within the interval A, let's say z, such that x < z < y. Since z lies between x and y, it should also be part of the open interval A.
6. Now, consider the output of the function f(z). Since f(x) = f(y), according to our assumption, it implies f(z) = f(x) = f(y).
7. However, this contradicts the fact that f(A) ≠ f(x), f(y) as stated in the original problem. We have found a value z within the open interval A for which f(z) is equal to f(x) and f(y), which goes against our assumption.
8. This contradiction negates our initial assumption and proves that there must exist an open interval A=(x,y), x ∈ B, and a function f:A --> B such that f(A) ≠ f(x), f(y).
By following these steps, you can prove the statement.