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Math Help - continuous functions over discrete space

  1. #1
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    continuous functions over discrete space

    Let (X,d) be a discrete metric space i.e

    d(x,y)=0 ,if x=y and d(x,y)=1 if x\neq y.

    Let (Y,ρ) be any metric space

    Prove that any function ,f from (X,d) to (Y,ρ) is continuous over X
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  2. #2
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    Quote Originally Posted by xalk View Post
    Let (X,d) be a discrete metric space i.e

    d(x,y)=0 ,if x=y and d(x,y)=1 if x\neq y.

    Let (Y,ρ) be any metric space

    Prove that any function ,f from (X,d) to (Y,ρ) is continuous over X
    let x_n be any sequence converging to x in X i.e.

    x_n \to x Using the sequential char of continuity

    We need to show that f(x_n) \to f(x) in Y

    But since we are in a discreate space x_n \to x
    only if there exists an N such that for all n > N x_n is constant(why?) and x_n=x

    But then for n>N \rho(f(x_n),f(x))=0
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  3. #3
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    Here's another approach:

     f: (X,d) \to (Y, \rho) is continuous at  x \in X if  \forall \varepsilon > 0, \exists \delta >0 such that  d(x,y)< \delta \Rightarrow \rho (f(x), f(y)) < \varepsilon .

    For any  \varepsilon >0 , we need to let  \delta be something such that the definition above is satisfied. Let's consider the two cases:

    (1)  0< \varepsilon < 1 , or
    (2)  1 \leq \varepsilon .

    In the first case, let's let  \delta = \varepsilon . Then  0< \delta < 1 . Since we're in a discrete metric space, whenever  d(x,y) < \delta , then  d(x,y) has to be 0 (because it's either 0 or 1, and by our choice of  \delta we've ruled-out the possibility of it being 1). BUT this means that  x=y (that is, the only time the distance between two points is zero is when the two points are, in fact, the same point). By definition of  f being a function, it must also be true that  f(x)=f(y) . This means, of course, that  \rho (f(x), f(y)) =0 . Since \varepsilon is strictly greater than 0, it is clearly true that  \rho (f(x),f(y)) < \varepsilon . Woohoo! The definition is satisfied!

    The second case is easier, and I'll leave the details of that one up to you if you choose to go this route. You're going to let  \delta = \varepsilon again, and using the defining property of a discrete metric space you'll see that the definition of continuity is satisfied pretty much automatically.
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    Quote Originally Posted by mylestone View Post
    Here's another approach:

     f: (X,d) \to (Y, \rho) is continuous at  x \in X if  \forall \varepsilon > 0, \exists \delta >0 such that  d(x,y)< \delta \Rightarrow \rho (f(x), f(y)) < \varepsilon .

    For any  \varepsilon >0 , we need to let  \delta be something such that the definition above is satisfied. Let's consider the two cases:

    (1)  0< \varepsilon < 1 , or
    (2)  1 \leq \varepsilon .

    In the first case, let's let  \delta = \varepsilon . Then  0< \delta < 1 . Since we're in a discrete metric space, whenever  d(x,y) < \delta , then  d(x,y) has to be 0 (because it's either 0 or 1, and by our choice of  \delta we've ruled-out the possibility of it being 1). BUT this means that  x=y (that is, the only time the distance between two points is zero is when the two points are, in fact, the same point). By definition of  f being a function, it must also be true that  f(x)=f(y) . This means, of course, that  \rho (f(x), f(y)) =0 . Since \varepsilon is strictly greater than 0, it is clearly true that  \rho (f(x),f(y)) < \varepsilon . Woohoo! The definition is satisfied!

    The second case is easier, and I'll leave the details of that one up to you if you choose to go this route. You're going to let  \delta = \varepsilon again, and using the defining property of a discrete metric space you'll see that the definition of continuity is satisfied pretty much automatically.
    THANK YOU .
    BUT considering the 2nd case,where \epsilon\geq 1 ,whether we put \delta =\epsilon,\delta<\epsilon,\delta>\epsilon,
    if we have d(x,y)<δ ,then the two cases :d(x,y)=0 ,d(x,y)=1 are both satisfied.
    And for the case d(x,y)=0 we can have ρ(f(x),f(y))<ε,
    but for the case d(x,y) =1,how can we have ρ(f(x),f(y))<ε??
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    Sorry, I didn't quite think that through. I guess you can just consider  \varepsilon = 1 and  \varepsilon > 1 separately. If  \varepsilon =1 , delta can be anything strictly between 0 and 1 and other than that, the same argument provided for (1) in my first reply should still work. If  \varepsilon > 1 , then  \rho (f(x),f(y)) \leq 1 < \varepsilon, \forall f(x), f(y) so  \delta = \varepsilon should be fine.

    That should do it, but I'm on way too much coffee and rushing out the door so if you find that I've botched it again, please let me know--thanks!
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  6. #6
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    Quote Originally Posted by mylestone View Post
    Sorry, I didn't quite think that through. I guess you can just consider  \varepsilon = 1 and  \varepsilon > 1 separately. If  \varepsilon =1 , delta can be anything strictly between 0 and 1 and other than that, the same argument provided for (1) in my first reply should still work. If  \varepsilon > 1 , then  \rho (f(x),f(y)) \leq 1 < \varepsilon, \forall f(x), f(y) so  \delta = \varepsilon should be fine.

    That should do it, but I'm on way too much coffee and rushing out the door so if you find that I've botched it again, please let me know--thanks!
    .

    The problem is with ε>1.

    Ιn this case even if we put δ=ε, we cannot prove that:

    d(x,y)<δ implies that ρ(f(x),f(y))<ε.Because in this case d(x,y) can be 0 and 1.

    If it is zero o.k then x=y and ρ(f(x).f(y)) =0<ε,but if it is 1,how can we prove that:

    ρ(f(x),f(y))< ε ??
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  7. #7
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    Quote Originally Posted by xalk View Post
    .

    The problem is with ε>1.

    Ιn this case even if we put δ=ε, we cannot prove that:

    d(x,y)<δ implies that ρ(f(x),f(y))<ε.Because in this case d(x,y) can be 0 and 1.

    If it is zero o.k then x=y and ρ(f(x).f(y)) =0<ε,but if it is 1,how can we prove that:

    ρ(f(x),f(y))< ε ??
    But in N with the discrete metric,  \rho (f(x),f(y)) must be less than or equal to 1 (being either 1 or 0). So if  \varepsilon >1 , then it is always the case that  \rho (f(x),f(y)) \leq 1 < \varepsilon . Epsilon is strictly greater than 1, and the distance between any two points of Y (in particular, any two points of the range of f) is at most--at most!--1. Does that clarify? Remember that Y has the discrete metric as well as X.  \delta can be 57,628,629,395,692,569,723,596,852 in this case if you want--the critical pieces of information here are that the distance between points in Y will never be more than one, and that epsilon is strictly greater than 1.

    Hope that helps, but if not maybe someone else can provide a better explanation (sometimes I'm not the clearest-headed feller).
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    Quote Originally Posted by mylestone View Post
    Remember that Y has the discrete metric as well as X.
    .

    No,no,i am sorry you misread my post.

    (Y,ρ) is any metric space.
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  9. #9
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    In a discrete metric space, any subset is clopen. Thus the inverse image
    f^{-1}(A) of any set A (subset of Y) is clopen in X. So, f is continuous.
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  10. #10
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    Quote Originally Posted by xalk View Post
    .

    No,no,i am sorry you misread my post.

    (Y,ρ) is any metric space.
    HAHAAAAA--wow, that's hilarious! Funny thing is I looked back on the original post...so I misread it twice!! Man, I do this way too often. I swear I'm literate.

    Let's try this again.

    So we're good when  \varepsilon \leq 1 and we want to ensure that when  \varepsilon >1 we can still find a  \delta that will satisfy continuity of f. Let  0< \delta \leq 1 . Then the same comment about  d(x,y) < \delta \Rightarrow x=y \Rightarrow f(x) = f(y) \Rightarrow \rho (f(x),f(y)) =0 < \varepsilon is relevant and continuity is satisfied. You don't have to ever worry about what happens when  \rho (f(x),f(y)) \geq 1 because it'll never happen as long as you pick delta appropriately.

    Alternatively, we can junk epsilons and deltas entirely if you've shown that f is continuous if and only if the preimage of an open set in Y is an open set in X. By this method, you can take an arbitrary open set in Y and no matter what it's preimage under any function is, the preimage must be open in X because every subset of a discrete metric space is open. So any function f from a discrete metric space into any other metric space is continuous.

    Of course, having carried a misreading this far I wish I would have advised that route to begin with. Apologies for my misunderstanding--thanks for playing!
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  11. #11
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    So for any ε>0,we choose δ:

    0<\delta\leq 1
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