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Question about a function of sets

  • #1
Let a function ##f:X \to X## be defined.
Let A and B be sets such that ##A \subseteq X## and ##B \subseteq X##.
Then which of the following are correct ?
a) ##f(A \cup B) = f(A) \cup f(B)##
b) ##f(A \cap B) = f(A) \cap f(B)##
c) ##f^{-1}(A \cup B) = f^{-1}(A) \cup f^{-1}(B)##
d) ##f^{-1}(A \cap B) = f^{-1}(A) \cap f^{-1}(B)##

My attempt-
For option a, let an element ##x \in f(A \cup B)##
##\Leftrightarrow## ##f^{-1}(x) \in A \cup B##
##\Leftrightarrow## ##f^{-1} (x)\in A## or ##f^{-1}(x) \in B##
##\Leftrightarrow## ##x \in f(A)## or ##x \in f(B)##
##\Leftrightarrow## ##x \in f(A) \cup f(B)##
##\Rightarrow## ##f(A \cup B) = f(A) \cup f(B)##

A similiar analogy can be applied to options c and d as well.
However, option b doesn't seem to fit into this argument. Even though this approach seems rationale, a counter example can be given to disprove option b. It goes as follows :
Let ##X=\left\{1,2,3 \right\}##, ##A=\left\{1 \right\}## and ##B=\left\{2 \right\}##
Let function ##f## be defined as ##f(1)=3## and ##f(2)=3##
Clearly ##A \cap B = \phi## and hence ##f(A \cap B)## becomes undefined.
Therefore disproving option b.
But option d is correct even though option b is incorrect. Can somebody clarify this for me ?
In simple terms, if option a is right then why not option b (surely there must be some flaw in the above proof when applied for option b but what is it) ? And since option b is incorrect how can option d be correct ?

NOTE: The above question appeared in an exam and the correct answers are options a,c,d.
 
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Answers and Replies

  • #2
Math_QED
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Let a function ##f:X \to X## be defined.
Let A and B be sets such that ##A \subseteq X## and ##B \subseteq X##.
Then which of the following are correct ?
a) ##f(A \cup B) = f(A) \cup f(B)##
b) ##f(A \cap B) = f(A) \cap f(B)##
c) ##f^{-1}(A \cup B) = f^{-1}(A) \cup f^{-1}(B)##
d) ##f^{-1}(A \cap B) = f^{-1}(A) \cap f^{-1}(B)##

My attempt-
For option a, let an element ##x \in f(A \cup B)##
##\Leftrightarrow## ##f^{-1}(x) \in A \cup B##
##\Leftrightarrow## ##f^{-1} (x)\in A## or ##f^{-1}(x) \in B##
##\Leftrightarrow## ##x \in f(A)## or ##x \in f(B)##
##\Leftrightarrow## ##x \in f(A) \cup f(B)##
##\Rightarrow## ##f(A \cup B) = f(A) \cup f(B)##

A similiar analogy can be applied to options c and d as well.
However, option b doesn't seem to fit into this argument. Even though this approach seems rationale, a counter example can be given to disprove option b. It goes as follows :
Let ##X=\left\{1,2,3 \right\}##, ##A=\left\{1 \right\}## and ##B=\left\{2 \right\}##
Let function ##f## be defined as ##f(1)=3## and ##f(2)=3##
Clearly ##A \cap B = \phi## and hence ##f(A \cap B)## becomes undefined.
Therefore disproving option b.
But option d is correct even though option b is incorrect. Can somebody clarify this for me ?
In simple terms, if option a is right then why not option b (surely there must be some flaw in the above proof when applied for option b but what is it) ? And since option b is incorrect how can option d be correct ?

NOTE: The above question appeared in an exam and the correct answers are options a,c,d.
What you did is correct, besides one detail. ##A \cap B = \emptyset## means that ##f( A \cap B) = \emptyset##. The latter set is not "undefined".

As for why (b) isn't correct and (d) is, this is just because preimage and image are very different concepts. For example, the image of a singelton will be a singelton, while the preimage of a singelton can be the entire domain (if the function is constant).

Preimages behave better than images with respect to set operations, as this exercice shows.
 
  • #3
PeroK
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If ##f^{-1}## exists, then ##f## (and ##f^{-1}##) must be one-to-one, which rules out your counterexample.

PS except, of course, ##f^{-1}## means a preimage in this context.
 
  • #4
Stephen Tashi
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Let a function ##f:X \to X## be defined.
How is the notation "##f:X \to X##" defined in your course materials? Does this signify that ##X## is the domain of ##f## ? - or does it include the possibility that the domain of ##f## is a proper subset of ##X##?

If it includes the possibility that the domain of ##f## can be proper subset of ##X## then can we say an empty set of ordered pairs is a function "##X \to X##" ?
 
  • #5
stevendaryl
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How is the notation "##f:X \to X##" defined in your course materials? Does this signify that ##X## is the domain of ##f## ? - or does it include the possibility that the domain of ##f## is a proper subset of ##X##?

If it includes the possibility that the domain of ##f## can be proper subset of ##X## then can we say an empty set of ordered pairs is a function "##X \to X##" ?
I don't know what the course's definition is, but I would say that ##f## is a function of signature ##X \to X##, then it means that the domain of ##f## is a superset of ##X## (you can further restrict it to say that the domain of ##f## is exactly ##X##, but for most purposes, it doesn't matter whether ##f## is defined for a larger domain that ##X## or not). If you wanted to allow the possibility that ##f## has a domain that may be smaller than ##X##, I would call that a "partial function" on ##X##.
 
  • #6
WWGD
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As a somewhat-trivial result that helps me with these arguments ( hopefully to you too) , all nice properties hold when you have a bijection, so, in a sense, the results you need to prove are somewhat of a measure of how non-bijective a function is.
 
  • #7
Yeah, I got it.
Thanks
 

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