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Math Help - Quick Continuity question

  1. #1
    Junior Member
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    Quick Continuity question

    Prove that if g:R->R is continuous at a then f(x,y)=g(x) is continuous at (a,b) \forall b \in R

    So we know
    \forall  e>0 \exists  d>0 s.t. \forall x \in R where |x-a|<d we have |g(x) - g(a)|<e
    So I've said as  \forall  b \in R g(x)=f(x,y) & g(a)=f(a,b), these can be substituted in giving the expression we need except for the condition that [(x-a)^2 + (y-b)^2]^\frac{1}{2}<d
    This seems to be an incorrect cheat though, am I along the right lines or not?
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  2. #2
    Senior Member
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    The order of your quantifiers is wrong. In particular, \forall  e>0 \exists  d>0 s.t. \forall x \in R implies uniform continuity, not continuity...

    You seem to have the right answer though; you can use the same epsilon and delta.
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  3. #3
    Super Member Failure's Avatar
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    Quote Originally Posted by kevinlightman View Post
    Prove that if g:R->R is continuous at a then f(x,y)=g(x) is continuous at (a,b) \forall b \in R

    So we know
    \forall e>0 \exists d>0 s.t. \forall x \in R where |x-a|<d we have |g(x) - g(a)|<e
    So I've said as  \forall b \in R g(x)=f(x,y) & g(a)=f(a,b), these can be substituted in giving the expression we need except for the condition that [(x-a)^2 + (y-b)^2]^\frac{1}{2}<d
    This seems to be an incorrect cheat though, am I along the right lines or not?
    Why should that move be an incorrect cheat? And, yes, you are on the right lines. What you need is simply that from \sqrt{(x-a)^2+(y-b)^2} < \delta it follows that |x-a|<\delta, which is true: |x-a| = \sqrt{(x-a)^2}\leq \sqrt{(x-a)^2+(y-b)^2} <\delta.
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