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Prove that if X is countable and a is in X then X \{a} is countable.

  1. Mar 8, 2012 #1
    Hi,

    Can someone please help me with this problem.
    Prove that if X is countable and a is in X then X \{a} is countable.

    this is what I have so far, and I am not sure if its correct:

    Every non-empty set of natural numbers has a least number. Since X is not a finite set, X must be an infinite set and thus X is nonempty in N. Suppose a in X is the least element. Now consider X\{a} Since X is infinite, X\{a} is an infinite subset of N. Then there is a least element a2 in X\{a}....

    thanks
     
  2. jcsd
  3. Mar 8, 2012 #2
    I would go about it this way: If X is countable, then there exists a subset of the natural numbers, say S, such that the cardinality of set S, call it r, is equal to the cardinality of set X. The set {a} has a cardinality of 1, so the cardinality of X\{a} is equal to r - 1 = n (some natural number, since we know X is nonempty). Then there exists a set S' such that its cardinality is equal to n, and therefore we have that X\{a} is countable.

    Perhaps I have left out some important information; I am not as smart as some of the people on here. I am only trying to help. Hopefully someone can support or refute my ideas here.
     
  4. Mar 8, 2012 #3
    A subset of countable set is countable. Since
    [tex]X\setminus\{a\}\subseteq X[/tex]

    we have that [itex]X\setminus\{a\}[/itex] is countable. So try proving first that any subset of a countable set is countable
     
  5. Mar 9, 2012 #4
    so,is this correct

    let x be countable set and x\{a} is subset of X

    if x\{a} is finite it is countable

    assume it is infinite then the function f : x|{a} ----> x defined by f(x0=x is 1-1 therefore x\{a} is countable


    Also, can someone help me to prove or give a counter example is sum ai and sum bi are convergent series with non-negative terms then sum aibi converges
     
  6. Mar 9, 2012 #5
    Let A be a subset of a countable set B. By the definition of a countable set, there exists a one-to-one function f : B→N. Now restrict the domain of f to A to give a new function
    g : A→N. g is one-to-one too because f is. So g is a one-to-one function that maps A into N. Thus A is countable

    and start new thread for new problem
     
  7. Mar 9, 2012 #6
    thank you
     
  8. Mar 9, 2012 #7

    HallsofIvy

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    Staff Emeritus
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    You are assuming that X is a subset of N and that is not given. All you are given is that X is countable- the set of all integers is countable, the set of all rational numbers is countable, .... Of course, if X is countable, there exist a one-to-one correspondence between X and the natural numbers.
     
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