Why does f(f^-1(E)) $\subset$ E?

  • Thread starter IniquiTrance
  • Start date
In summary, the function f might not be surjective, meaning that there could be an x in E where f-1({x})=∅. In this case, f(f-1({x}))=∅ which is a proper subset of {x}. Similarly, if f is not bijective, there could be instances where f^{-1}(f(E))\supset E. This happens when the function is non-bijective, such as in the example of f(x)= x^2, where E=[-1, 4] and f(E)=[0, 4].
  • #1
IniquiTrance
190
0
Hey all,

How is it that [itex] f(f^{-1}(E))\subset E[/itex] for some induced metric [itex]E\subset Y[/itex], can be a proper subset of E, rather than E itself?

Thanks!
 
Physics news on Phys.org
  • #2
The function f might not be surjective. If it were not, there would be an x in E such that f-1({x})=∅. Then f(f-1({x}))=∅, which is a proper subset of {x}.
 
  • #3
A. Bahat said:
The function f might not be surjective. If it were not, there would be an x in E such that f-1({x})=∅. Then f(f-1({x}))=∅, which is a proper subset of {x}.

Thank you.
 
  • #4
One more question, say [itex]E\subset X[/itex] s.t. [itex]f^{-1}(f(E))\supset E[/itex]. How can this occur? Is this the case when a function is non-bijective, such as where f(a)=f(b) for [itex]a \neq b,a \in E, b\notin E [/itex]
 
Last edited:
  • #5
Let's take a specific example: [itex]f(x)= x^2[/itex] which is neither surjective ("onto" the real numbers) nor injective ("one to one").
Let E= [-1, 4]. Then [itex]f^{-1}(E)= [-2, 2][/itex] and [itex]f(f^{-1}(E))= [0, 4][/itex] which is a proper subset of E.

Let E= [0, 2]. Then [itex]f(E)= [0, 4][/itex] and [itex]f^{-1}(f(E))= [-2, 2][/itex] which contains E as a proper subsert.
 
  • #6
Thank you.
 

1. Why does f(f^-1(E)) $\subset$ E?

The reason why f(f^-1(E)) $\subset$ E is because the function f^-1(E) maps elements of the set E back to their original elements in the domain of f. When we apply the function f to these elements, it simply transforms them back to their original form, thus resulting in a subset of E.

2. How does f(f^-1(E)) $\subset$ E relate to the concept of inverse functions?

f(f^-1(E)) $\subset$ E is a direct result of the definition of an inverse function. The inverse function f^-1(E) is defined as the function that maps elements of E back to their original elements in the domain of f. Thus, when we apply f to these elements, it will result in a subset of E.

3. Can you provide an example to illustrate f(f^-1(E)) $\subset$ E?

Let's say we have a function f(x) = 2x, where x is any real number. The inverse function of this would be f^-1(x) = x/2. Now, let's take the set E = {2, 4, 6}. If we apply the inverse function f^-1 to this set, we get {1, 2, 3}. And when we apply the original function f to this set, we get {2, 4, 6}, which is a subset of our original set E.

4. Why is it important to understand the concept of f(f^-1(E)) $\subset$ E?

Understanding the concept of f(f^-1(E)) $\subset$ E is important because it helps us to understand the relationship between a function and its inverse. It also helps in solving equations involving inverse functions and in proving mathematical theorems related to functions.

5. Is f(f^-1(E)) $\subset$ E always true for any function f and set E?

Yes, f(f^-1(E)) $\subset$ E is always true for any function f and set E. This is because the definition of an inverse function guarantees that the elements of the image of f^-1(E) will always be a subset of the original set E.

Similar threads

  • Differential Geometry
Replies
7
Views
3K
  • Differential Geometry
Replies
20
Views
2K
  • Topology and Analysis
Replies
14
Views
342
  • Calculus and Beyond Homework Help
Replies
2
Views
830
  • Differential Geometry
Replies
9
Views
2K
  • Set Theory, Logic, Probability, Statistics
Replies
1
Views
1K
Replies
4
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
458
Replies
3
Views
578
Replies
2
Views
255
Back
Top