Math question re continuity
November 19, 2010 9:26 AM Subscribe
A function is continuous over a closed interval [a,b]
***(1) if it is continuous over the open interval (a,b) and
***(2) the lim as x approaches a from the right = f(a) and the limit as x approaches b from the left =f(b).
Can someone describe a function over a closed interval that fails continuity by either or both of condition (2)?
Best answer: Let f(x) = 0 everywhere, except f(0) = 1. Then f(x) is continuous on the open interval (0,1), but the limit of f as x goes to 0 is not equal to f(0).
Or am I misunderstanding what you are asking?
posted by AlexanderPetros at 9:34 AM on November 19, 2010
Or am I misunderstanding what you are asking?
posted by AlexanderPetros at 9:34 AM on November 19, 2010
There are many examples. Pick any of your favorite continuous functions, but just redefine them at the endpoints.
Such as, f(x) = x2, except at 0, we have f(0) = 1. This function is not continuous on the closed interval [0,1], for instance.
posted by King Bee at 9:38 AM on November 19, 2010
Such as, f(x) = x2, except at 0, we have f(0) = 1. This function is not continuous on the closed interval [0,1], for instance.
posted by King Bee at 9:38 AM on November 19, 2010
Response by poster: Let f(x) = 0 everywhere, except f(0) = 1.
Duh!! Thank you.
posted by Neiltupper at 9:56 AM on November 19, 2010
Duh!! Thank you.
posted by Neiltupper at 9:56 AM on November 19, 2010
Any function with a vertical asymptote - the lim as f->a is infinity, but f(a) isn't "infinity" but rather "does not exist".
Until you get into really weird stuff, most functions that fail this are degenerate. However, certain calculus principles rely on a function being continuous on a closed interval, which is why they're even bringing it up, probably.
posted by notsnot at 9:58 AM on November 19, 2010
Until you get into really weird stuff, most functions that fail this are degenerate. However, certain calculus principles rely on a function being continuous on a closed interval, which is why they're even bringing it up, probably.
posted by notsnot at 9:58 AM on November 19, 2010
More to the point, or rather to give an only moderately degenerate answer that doesn't rely on funny definitions, look at the funcion (x2-3x+2)/(x-1). the numerator factors to (x-1)(x-2), which cancels with the x-1 on teh bottom, so the function looks like y=x-2 EXCEPT at x=1. At that point, in the original function definition (not the "y=x-2" you're using for simplicity), you're dividing by zero. So there's a hole there. Not an asymptote, just a hole. If you were to attempt to determine continuity on the closed interval [0,1] it would fail. Open interval, no problem.
posted by notsnot at 10:03 AM on November 19, 2010
posted by notsnot at 10:03 AM on November 19, 2010
look at the funcion (x2-3x+2)/(x-1). the numerator factors to (x-1)(x-2), which cancels with the x-1 on teh bottom, so the function looks like y=x-2 EXCEPT at x=1.
Not only is this function not continuous on [0,1], it's not even defined at every point in the interval. So, this kind of example fails the OP's (2) in spectacular fashion.
posted by King Bee at 11:38 AM on November 19, 2010
Not only is this function not continuous on [0,1], it's not even defined at every point in the interval. So, this kind of example fails the OP's (2) in spectacular fashion.
posted by King Bee at 11:38 AM on November 19, 2010
Not only is this function not continuous on [0,1], it's not even defined at every point in the interval. So, this kind of example fails the OP's (2) in spectacular fashion.
actually, it fails the condition in a really mundane way: it's not a function on the domain [0,1]. The OP's original definition is missing the fine print which says the function must be defined on the closed interval. However it's possible to extend it to a continuous function on the domain [0,1] by 'pasting' f(1) = -1.
Any function with a vertical asymptote - the lim as f->a is infinity, but f(a) isn't "infinity" but rather "does not exist".
technically speaking, this is not an example of a discontinuous function on the domain [a,b] because it isn't a well-defined function on that domain. This point is almost always lost in Calc 101 discussions of real function theory. it is however, an example of a function on (a,b) which *cannot* be extended to a continuous function on [a,b] like the previous example was...
it's a subtle point which is probably not worth considering at this level but I feel compelled anyway.
posted by ennui.bz at 11:53 AM on November 19, 2010
actually, it fails the condition in a really mundane way: it's not a function on the domain [0,1]. The OP's original definition is missing the fine print which says the function must be defined on the closed interval. However it's possible to extend it to a continuous function on the domain [0,1] by 'pasting' f(1) = -1.
Any function with a vertical asymptote - the lim as f->a is infinity, but f(a) isn't "infinity" but rather "does not exist".
technically speaking, this is not an example of a discontinuous function on the domain [a,b] because it isn't a well-defined function on that domain. This point is almost always lost in Calc 101 discussions of real function theory. it is however, an example of a function on (a,b) which *cannot* be extended to a continuous function on [a,b] like the previous example was...
it's a subtle point which is probably not worth considering at this level but I feel compelled anyway.
posted by ennui.bz at 11:53 AM on November 19, 2010
I agree, it should be mentioned that the definition of continuity on a set includes that the function must be at least defined on that set.
Don't you think that a function which is not even defined on a set "spectacularly fails" to be continuous there? For instance, the square root function is "spectacularly" not continuous at any negative real number.
It's Friday, and this mathematics instructor is trying to have some fun. =)
posted by King Bee at 11:57 AM on November 19, 2010
Don't you think that a function which is not even defined on a set "spectacularly fails" to be continuous there? For instance, the square root function is "spectacularly" not continuous at any negative real number.
It's Friday, and this mathematics instructor is trying to have some fun. =)
posted by King Bee at 11:57 AM on November 19, 2010
I think the Treaty of Westphalia even more spectacularly fails to be continuous on [0,1]. ;)
Slightly more relatedly, I've always loved the everywhere continuous but nowhere differentiable Weierstrass function. Fractally pathological!
posted by kmz at 12:55 PM on November 19, 2010 [1 favorite]
Slightly more relatedly, I've always loved the everywhere continuous but nowhere differentiable Weierstrass function. Fractally pathological!
posted by kmz at 12:55 PM on November 19, 2010 [1 favorite]
My favourite pathological mathematical function: f(x) = x if x is a rational number, and f(x) = 0 otherwise. This function is continuous at x = 0, but for no other x.
posted by Johnny Assay at 1:46 PM on November 19, 2010
posted by Johnny Assay at 1:46 PM on November 19, 2010
kmz, you have a good point!
My favorite pathological function is the Cantor function It is continuous, but not absolutely. It is differentiable almost everywhere, and its derivative is 0 almost everywhere. It maps the Cantor Set onto [0,1]. OMGWTFBBQ
posted by King Bee at 5:34 PM on November 19, 2010
My favorite pathological function is the Cantor function It is continuous, but not absolutely. It is differentiable almost everywhere, and its derivative is 0 almost everywhere. It maps the Cantor Set onto [0,1]. OMGWTFBBQ
posted by King Bee at 5:34 PM on November 19, 2010
This thread is closed to new comments.
f(x) = 1 if x is in (a,b)
f(x) = 0 if x = a or x = b
posted by alk at 9:32 AM on November 19, 2010 [1 favorite]