Set of functions that is eventually zero

AI Thread Summary
The discussion clarifies the concept of functions f: N → {0,1} that are "eventually zero," meaning they eventually consist solely of zeros after a certain point. Participants confirm that these functions can be interpreted as binary sequences where, after a finite number of 1's, the sequence continues with 0's indefinitely. The correct way to visualize this is by reading the sequence from left to right, not right to left. Thus, a binary sequence is considered eventually zero if it contains only a limited number of 1's. Understanding this concept is essential for solving related homework problems.
Marioqwe
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Usually, in homework problems, I come across something like, "Let F be the set of all functions f:\mathbf{N}\rightarrow\{0,1\} that are eventually zero."

But I don't really understand what is meant by that. Is it right to think about it as the set of binary numbers? If I take each f to be a sequence of 0's and 1's and I read them from right to left then they are eventually zero right? I'm not sure this is the right way of thinking about this.
 
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Hi Marioqwe! :smile:
Marioqwe said:
… If I take each f to be a sequence of 0's and 1's and I read them from right to left then they are eventually zero right?

yes

an f:N -> {0,1} is a sequence of 0s and 1s

for example, 110110100100000000000000…

if it ends with all 0s after some time ("zero recurring"), then it is eventually zero :wink:
 
Marioqwe said:
Usually, in homework problems, I come across something like, "Let F be the set of all functions f:\mathbf{N}\rightarrow\{0,1\} that are eventually zero."

But I don't really understand what is meant by that. Is it right to think about it as the set of binary numbers? If I take each f to be a sequence of 0's and 1's and I read them from right to left then they are eventually zero right? I'm not sure this is the right way of thinking about this.

Did you mean left to right? That's how to think of this. Another way to say it is that a binary sequence is eventually zero if it contains only finitely many 1's.
 
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