A Function is Continuous at an Isloated Point
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Isolated points and continuity
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Homework Statement
Let f : A --> R be a function, and let c in A be an isolated point of A. Prove that f
is continuous at c
Homework Equations
The Attempt at a Solution
I'm kind of confused by this problem.... if c is an isolated point, then the limit doesn't exist. So I can't really use the fact that a function is continuous at c if for all epsilon>0 there exists a delta>0 such that whenever |x-c|<delta, it follows that |f(x)-f(c)|<epsilon.
Any hints would be great!
Answers and Replies
I am not sure how you are choosing delta.... are you saying that you choose delta to be equal to the delta satisfying (c-delta, c+delta) containing no other points of A but c?
Yes. That delta exists by the fact that c is an isolated point. Now just go through the epsilon-delta steps of proving f is continuous at c. Using that delta, what does [itex]|x-c|<\delta[/itex] imply if x is in A?
Because I'm trying to show that
|f(x)-L|< epsilon, but there is no limit so that is why I'm getting confused....
Would it just mean that x-c is in a as well?
Because I'm trying to show that
|f(x)-L|< epsilon, but there is no limit so that is why I'm getting confused....
Well your L is f(c). So you are trying to show that for x in A and any [itex]\varepsilon>0[/itex], there is a [itex]\delta>0[/itex] such that [itex]|x-c|<\delta[/itex] implies that [itex]|f(x)-f(c)|<\varepsilon[/itex]. But if x is in A and [itex]|x-c|<\delta[/itex], where the delta is the one described above, then x can only be one point! And that point is ...?
Well it wouldn't be true for x=c. Because |x-c| would be equal to zero, and that contradicts the statement that delta is greater than zero.
Sorry, but that doesn't make sense.
How does any of that answer the question, can the statement "a function is continuous at c if for all epsilon>0 there exists a delta>0 such that whenever |x-c|<delta, it follows that |f(x)-f(c)|<epsilon" be untrue?
Choose any ε … for example choose ε = 2009 …
Choose any ε … for example choose ε = 2009 …
I understand that I can find a delta for any specific epsilon, but how am I supposed to phrase that in my proof for an arbitrary epsilon?
Let c be an isolated point of A. Then there exists a [itex]\delta>0[/itex] such that the interval [itex](c-\delta,c+\delta)[/itex] contains no other points of A besides c.
Let x be in A and [itex]\varepsilon>0[/itex]. Then [itex]|x-c|<\delta[/itex] implies that x is in the interval [itex](c-\delta,c+\delta)[/itex]. This means x=c because there are no others points of A in that interval. This implies that [itex]|f(x)-f(c)|=|f(c)-f(c)|=0<\varepsilon[/itex] for absolutely any [itex]\varepsilon>0[/itex] you choose. Therefore, f is continuous at c.
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Source: https://www.physicsforums.com/threads/isolated-points-and-continuity.350877/
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