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Some set proofs

  • Thread starter Geekster
  • Start date
Disclaimer: I might have some problems getting my LaTeX code to work properly so please bear with me while I figure out how to properly use the forum software.

1. Homework Statement
The exercise is to prove the following statements.

Suppose that [itex]f:X \rightarrow Y[/itex], the following statement is true.
If [itex]\{G_{\alpha} : \alpha \in A\}[/itex] is an indexed family of subsets of Y, then [itex]f^{-1}(\bigcup_{\alpha \in A} G_\alpha) =\bigcup_{\alpha \in A} f^{-1}( G_\alpha) [/itex].

2. Homework Equations

The relevant information in this case is the definitions. In this case I need to know what the definition of an inverse function is.

DEF: Suppose [itex]f:X \rightarrow Y[/itex] and [itex] A \subset Y[/itex]. [itex]f^{-1}(A) = \{x \in X: f(x) \in A\}[/itex]

3. The Attempt at a Solution

The solution I've been looking thus far is a point wise argument.

Choose [itex]t \in f^{-1}(\bigcup_{\alpha \in A} G_\alpha)[/itex]. So by definition we know
[itex]t \in \{x \in X: f(x) \in (\bigcup_{\alpha \in A} G_\alpha)\}[/itex]. And here is where I'm kind of stuck. I need to some how get my chosen element to be in the other set, therefore making the one set a subset of the other. Then complete the converse of the argument to finish the proof.

Any ideas on where I should be looking, or what I should be thinking about here?
Last edited:

Answers and Replies

Homework Helper
[itex]f(x) \in \cup_{\alpha \in A} G_\alpha[/itex] is equivalent to "there is some [itex]\alpha \in A[/itex] such that [itex]f(x) \in G_\alpha[/itex]". Also, you don't have to worry about doing the converse if every step you take is an equivalence.

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