http://www.bing.com:80/search?q=Continuous+Data+Collection+ProblemsSearch resultshttp://www.bing.com:80/s/a/rsslogo.gifhttp://www.bing.com:80/search?q=Continuous+Data+Collection+ProblemsCopyright © 2026 Microsoft. All rights reserved. These XML results may not be used, reproduced or transmitted in any manner or for any purpose other than rendering Bing results within an RSS aggregator for your personal, non-commercial use. Any other use of these results requires express written permission from Microsoft Corporation. By accessing this web page or using these results in any manner whatsoever, you agree to be bound by the foregoing restrictions.https://math.stackexchange.com/questions/5114829/continuous-vs-discrete-variablesBoth discrete and continuous variables generally do have changing values—and a discrete variable can vary continuously with time. I am quite aware that discrete variables are those values that you can count while continuous variables are those that you can measure such as weight or height.Sat, 28 Mar 2026 22:48:00 GMThttps://math.stackexchange.com/questions/5127388/proving-a-limit-of-a-measure-is-continuousI was trying to formalize some things about string motion in physics so I could answer more general questions about it and then I got to a point as to see the limit written below. I then asked myselfFri, 10 Apr 2026 13:00:00 GMThttps://math.stackexchange.com/questions/477/cardinality-of-set-of-real-continuous-functionsThe cardinality is at most that of the continuum because the set of real continuous functions injects into the sequence space $\mathbb R^N$ by mapping each continuous function to its values on all the rational points. Since the rational points are dense, this determines the function.Mon, 13 Apr 2026 14:29:00 GMThttps://math.stackexchange.com/questions/1182795/what-is-the-intuition-for-semi-continuous-functionsA function is continuous if the preimage of every open set is an open set. (This is the definition in topology and is the "right" definition in some sense.) The definitions you cite of semicontinuities claim that the preimages of certain open sets are open, but does not say so about all open sets. Note that $\ { \ {f \in \mathbb {R} \mid f > \alpha\} \mid \alpha \in \mathbb {R} \} \cup ...Sat, 11 Apr 2026 19:11:00 GMThttps://math.stackexchange.com/questions/254626/the-space-of-bounded-continuous-functions-is-not-separableThe space of bounded continuous functions is not separable Ask Question Asked 13 years, 4 months ago Modified 3 months agoSat, 11 Apr 2026 19:11:00 GMThttps://math.stackexchange.com/questions/3364/topological-properties-preserved-by-continuous-mapsYou'll find topological properties with indication of whether they are preserved by (various kinds of) continuous maps or not (such as open maps, closed maps, quotient maps, perfect maps, etc.). For mere continuous most things have been mentioned: simple covering properties (variations on compactness, connectedness, Lindelöf) and separability.Fri, 10 Apr 2026 13:00:00 GMThttps://math.stackexchange.com/questions/3719652/the-product-of-continuous-function-is-continuousThe product of continuous function is continuous. Ask Question Asked 5 years, 10 months ago Modified 1 year, 11 months agoSat, 11 Apr 2026 22:53:00 GMThttps://math.stackexchange.com/questions/4853516/can-a-discontinuous-function-have-a-continuous-derivativeCan a discontinuous function have a continuous derivative? Ask Question Asked 2 years, 2 months ago Modified 2 years, 2 months agoSun, 12 Apr 2026 19:03:00 GMThttps://math.stackexchange.com/questions/2825505/why-not-include-as-a-requirement-that-all-functions-must-be-continuous-to-be-difWe know that differentiable functions must be continuous, so we define the derivative to only be in terms of continuous functions. But then, the fact that differentiable functions are continuous is by definition, while it is being used to justify that very definition.Tue, 07 Apr 2026 17:21:00 GMThttps://math.stackexchange.com/questions/220733/is-a-bounded-and-continuous-function-uniformly-continuousThis function is continuous on $ [-1,1]$, so it is uniformly continuous there. A fortiori on $ (-1,1)$.Sun, 12 Apr 2026 01:16:00 GMT