3 ms·
I believe that the function you just defined is already continuous. It's not defined at 1 so checking its continuity doesn't make sense there, and at every othe
by GolDDranks 7y ago
I believe that the function you just defined is already continuous. It's not defined at 1 so checking its continuity doesn't make sense there, and at every other point it's clearly continuous because it's constant. Points that are arbitrarily close to 1 are still continuous because the intervals around 1 are open.
- Retric 7y agoIn calculus f(x) = x/x - 1 is not continuous at 1 based on the initial definitions. Moving to fixed-point arithmetic the idea of limits gets odd, but the function still needs a definition for that input.
- GolDDranks 7y agoI would say the continuity isn't just defined at that point... I very much agree that one can't affirmatively say that the function is continuous at that point. Btw. I guess you meant to say f(x) = (x - 1)/(x - 1). However, let's ignore that; the problem disappears if we define f(x ) = 1 at x = 1, and f(x) = 0 elsewhere. In standard analysis that function would be discontinuous at that point. I'm not sure about implications in the OP's kind of analysis. Would the existence of m(1) imply that there exist some smallest input that is larger than 1, that would make difference in output? Same for the largest input that is smaller than 1.
- tel 7y agoYou handle continuity in discrete spaces through topology. A function is continuous for a (really, two) given topology if the inverse images of open sets are open sets.