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The simplest gotcha I know is: is f(x) = 1/x piece-wise continuous?. This is calculus 1 level material and yet author's disagree significantly on this point, so

by tgb 6y ago

The simplest gotcha I know is: is f(x) = 1/x piece-wise continuous?. This is calculus 1 level material and yet author's disagree significantly on this point, sometimes without specifying it! Some say yes, others would require f to have finite left and right limits at every point. This mattered for a point of my thesis and my advisor was very unhappy with me calling these function piece-wise continuous.

gowld 6y ago
f doesn't even have either of a right or left limit at 0. f is maybe (piece-wise) continuous over what pseudo-domain? R or R\{0}? You could axiomitize that an infinite discontinuity is like an unbouned function as x->infinity, but then how would you avoid 1/x being regular continuous? I think you are claimokg that a set being incomplete "at the end" is different from a set having a hole in the middle -- that the question of continuity presumes connected sets. That's not standard but might be an appropriate assumption for the context of your paper. Anyway, arguing over terminology is boring unless it raises conceptual issues. The point is to communicate, which has at least 2 stakeholders. Clarify your terms and move on.
- Ericson2314 6y ago
  The idea is that what 1/x is is intuitively simple enough, so we should have some standard terminology for it. And saying left and right limits is -infinity or +infinity really also isn't that weird. I'm pretty sure other metric spaces can be likewise extended and end up with similar algebraic laws as the "extended real numbers". Again this isn't very profound, but is good for education and efficient communication, and so should be perused. Finally, it's interesting that measure spaces with infinite measure is already a thing that people. I would like to see more connections with metric and measure spaces; e.g. we can have an n-point metric which is defined using the measure of the (n-1)-simplex. Just as regular metric spaces have a "triangle inequality", 3-point metric spaces should have a "tetrahedron inequality", and n-point metric spaces should have a "(n-1)-simplex inequality". Again, this is not profound, but good for communication, and connections between definitions help one compress everything for better mental storage.
- tgb 6y ago
  I think you missed the point of the example, which is that people rarely clarify this because to the author their definition seems obviously correct. I made it 90% through a PhD without having considered that there could be more than one possible meaning for "piece-wise continuous". The domain is R in this case. The less-restrictive definition would be that f is piece-wise continuous if there are a discrete set of points .. < x_0 < x_1 < ... with f continuous on each interval (x_i, x_{i+1}). The alternative definition is that, plus requiring that the restriction of f to those intervals have limits at the endpoint. For piecewise smooth function, there's an even larger variety of possible meanings, yet it's often stated without clarification.
  - thaumasiotes 6y ago
    > The less-restrictive definition would be that f is piece-wise continuous if there are a discrete set of points .. < x_0 < x_1 < ... with f continuous on each interval (x_i, x_{i+1}). The alternative definition is that, plus requiring that the restriction of f to those intervals have limits at the endpoint. This sounds kind of surreal; putting them in shorter terms, we have two rival definitions for "piecewise continuous": 1. A function f is piecewise continuous if there exist one or more intervals over which f is continuous. 2. A function f is piecewise continuous if there exist one or more intervals over which f is (1) continuous, and (2) bounded. I agree that it sounds obvious which of those is more appropriate as a definition of "piecewise continuous"... (It also worries me that the definition you give requires the intervals to be adjacent, but doesn't require that more than one interval exist. A function that is continuous over (-2, -1) and also over (1, 2), but not anywhere else, would meet this definition, but you wouldn't be able to include both of those intervals of continuity in the set of endpoints. I would prefer to either have two sets of endpoints, such that we end up saying f is continuous over (x_i, y_i), (x_{i+1}, y_{i+1}), etc. (if we want to allow for intervals of discontinuity), or to say that the intervals (-inf, x_0) and (x_n, +inf) also count (if we don't). However, if we take that second approach, and we go with the definition of piecewise continuity that requires the function be bounded over every interval, we've just defined functions like f(x) = x as being not piecewise continuous despite the fact that they are continuous.)
    - 6y ago
gus_massa 6y ago
I teach math in the first year of the university and it's not a good moment to discuss about the subtle details of topology and definitions. So every time someone ask me that, I reply "It is not continuous for all the real numbers"
- thaumasiotes 6y ago
  Well, obviously it's not continuous. How is that an answer to "is it piecewise continuous?"?
  - gus_massa 6y ago
    It is continuous in the natural domain[1] that is (-∞,0)∪(0,+∞). The problem is that the students of the first (or second) year of the university will then try to use the the Bolzano's theorem / Intermediate value theorem to prove that it has a zero in the interval [-1, 1]. So I must answer NO, but the problem is that usually in the question the domain is implicit, so it cause a lot of confusion. I think it's more clear the definition of "piecewise differentiable". I'm not sure what is the "official" definition of "piecewise continuous". The definition in Mathworld looks a little fuzzy. I'd probably request not an horrible behavior in the borders of the intervals that are glued, like in * sign(x) -> yes * 1/x -> no * sin(1/x) -> no One of the best keep secrets in math is that the definitions are somewhat arbitrary. [1] At least it is how we call it in Argentina, sometimes the names/definitions change in each country.
generationP 6y ago
I thought this was only an issue in K-12, as anyone in research math considers the domain and the target to be part of a function, and then the problem disappears: The function R \ {0} -> R, x |-> 1/x is not just piecewise continuous but continuous on-the-nose. The function R -> R, x |-> 1/x doesn't exist. The function R |-> some completion of R, x |-> 1/x is continuous or not depending on which completion you choose (the one with two infinities or the projective line). But I do recall a similar confusion happening with "piecewise linear" (the question is whether the pieces have to fit together).
- nicoburns 6y ago
  > I thought this was only an issue in K-12, as anyone in research math considers the domain and the target to be part of a function Anyone is research math does consider the domain and target to be part of the function. Alas, that doesn't mean that they'll actually write down which domain and target they have in mind.
- tgb 6y ago
  No you've missed the distinction. In all cases the domain and range are R (you can fill in a value at 0, it doesn't matter which). See the MathWorld page which leaves the definition intentionally ambiguous: "A function or curve is piecewise continuous if it is continuous on all but a finite number of points at which certain matching conditions are sometimes required." Emphasis added. https://mathworld.wolfram.com/PiecewiseContinuous.html https://mathworld.wolfram.com/PiecewiseContinuous.html
  - generationP 6y ago
    I don't get this. If you require f to be piecewise continuous outside of a finite set of points and to satisfy left limit = right limit at each of these points, then you just have a continuous function. Why another word for it?
    - tgb 6y ago
      The left and right limits are required to exist (and finite), but not necessarily to be equal to each other. So f(x) = 1/x and sin(1/x) are out but x/abs(x) is not.

by tgb 6y ago

gowld 6y ago

f doesn't even have either of a right or left limit at 0. f is maybe (piece-wise) continuous over what pseudo-domain? R or R\{0}? You could axiomitize that an infinite discontinuity is like an unbouned function as x->infinity, but then how would you avoid 1/x being regular continuous? I think you are claimokg that a set being incomplete "at the end" is different from a set having a hole in the middle -- that the question of continuity presumes connected sets. That's not standard but might be an appropriate assumption for the context of your paper. Anyway, arguing over terminology is boring unless it raises conceptual issues. The point is to communicate, which has at least 2 stakeholders. Clarify your terms and move on.

Ericson2314 6y ago
The idea is that what 1/x is is intuitively simple enough, so we should have some standard terminology for it. And saying left and right limits is -infinity or +infinity really also isn't that weird. I'm pretty sure other metric spaces can be likewise extended and end up with similar algebraic laws as the "extended real numbers". Again this isn't very profound, but is good for education and efficient communication, and so should be perused. Finally, it's interesting that measure spaces with infinite measure is already a thing that people. I would like to see more connections with metric and measure spaces; e.g. we can have an n-point metric which is defined using the measure of the (n-1)-simplex. Just as regular metric spaces have a "triangle inequality", 3-point metric spaces should have a "tetrahedron inequality", and n-point metric spaces should have a "(n-1)-simplex inequality". Again, this is not profound, but good for communication, and connections between definitions help one compress everything for better mental storage.
tgb 6y ago
I think you missed the point of the example, which is that people rarely clarify this because to the author their definition seems obviously correct. I made it 90% through a PhD without having considered that there could be more than one possible meaning for "piece-wise continuous". The domain is R in this case. The less-restrictive definition would be that f is piece-wise continuous if there are a discrete set of points .. < x_0 < x_1 < ... with f continuous on each interval (x_i, x_{i+1}). The alternative definition is that, plus requiring that the restriction of f to those intervals have limits at the endpoint. For piecewise smooth function, there's an even larger variety of possible meanings, yet it's often stated without clarification.
- thaumasiotes 6y ago
  > The less-restrictive definition would be that f is piece-wise continuous if there are a discrete set of points .. < x_0 < x_1 < ... with f continuous on each interval (x_i, x_{i+1}). The alternative definition is that, plus requiring that the restriction of f to those intervals have limits at the endpoint. This sounds kind of surreal; putting them in shorter terms, we have two rival definitions for "piecewise continuous": 1. A function f is piecewise continuous if there exist one or more intervals over which f is continuous. 2. A function f is piecewise continuous if there exist one or more intervals over which f is (1) continuous, and (2) bounded. I agree that it sounds obvious which of those is more appropriate as a definition of "piecewise continuous"... (It also worries me that the definition you give requires the intervals to be adjacent, but doesn't require that more than one interval exist. A function that is continuous over (-2, -1) and also over (1, 2), but not anywhere else, would meet this definition, but you wouldn't be able to include both of those intervals of continuity in the set of endpoints. I would prefer to either have two sets of endpoints, such that we end up saying f is continuous over (x_i, y_i), (x_{i+1}, y_{i+1}), etc. (if we want to allow for intervals of discontinuity), or to say that the intervals (-inf, x_0) and (x_n, +inf) also count (if we don't). However, if we take that second approach, and we go with the definition of piecewise continuity that requires the function be bounded over every interval, we've just defined functions like f(x) = x as being not piecewise continuous despite the fact that they are continuous.)
  - 6y ago

gus_massa 6y ago

I teach math in the first year of the university and it's not a good moment to discuss about the subtle details of topology and definitions. So every time someone ask me that, I reply "It is not continuous for all the real numbers"

thaumasiotes 6y ago
Well, obviously it's not continuous. How is that an answer to "is it piecewise continuous?"?
- gus_massa 6y ago
  It is continuous in the natural domain[1] that is (-∞,0)∪(0,+∞). The problem is that the students of the first (or second) year of the university will then try to use the the Bolzano's theorem / Intermediate value theorem to prove that it has a zero in the interval [-1, 1]. So I must answer NO, but the problem is that usually in the question the domain is implicit, so it cause a lot of confusion. I think it's more clear the definition of "piecewise differentiable". I'm not sure what is the "official" definition of "piecewise continuous". The definition in Mathworld looks a little fuzzy. I'd probably request not an horrible behavior in the borders of the intervals that are glued, like in * sign(x) -> yes * 1/x -> no * sin(1/x) -> no One of the best keep secrets in math is that the definitions are somewhat arbitrary. [1] At least it is how we call it in Argentina, sometimes the names/definitions change in each country.

generationP 6y ago

I thought this was only an issue in K-12, as anyone in research math considers the domain and the target to be part of a function, and then the problem disappears: The function R \ {0} -> R, x |-> 1/x is not just piecewise continuous but continuous on-the-nose. The function R -> R, x |-> 1/x doesn't exist. The function R |-> some completion of R, x |-> 1/x is continuous or not depending on which completion you choose (the one with two infinities or the projective line). But I do recall a similar confusion happening with "piecewise linear" (the question is whether the pieces have to fit together).

nicoburns 6y ago
> I thought this was only an issue in K-12, as anyone in research math considers the domain and the target to be part of a function Anyone is research math does consider the domain and target to be part of the function. Alas, that doesn't mean that they'll actually write down which domain and target they have in mind.
tgb 6y ago
No you've missed the distinction. In all cases the domain and range are R (you can fill in a value at 0, it doesn't matter which). See the MathWorld page which leaves the definition intentionally ambiguous: "A function or curve is piecewise continuous if it is continuous on all but a finite number of points at which certain matching conditions are sometimes required." Emphasis added. https://mathworld.wolfram.com/PiecewiseContinuous.html https://mathworld.wolfram.com/PiecewiseContinuous.html
- generationP 6y ago
  I don't get this. If you require f to be piecewise continuous outside of a finite set of points and to satisfy left limit = right limit at each of these points, then you just have a continuous function. Why another word for it?
  - tgb 6y ago
    The left and right limits are required to exist (and finite), but not necessarily to be equal to each other. So f(x) = 1/x and sin(1/x) are out but x/abs(x) is not.