Can a function be continuous on a composed interval?

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Discussion Overview

The discussion revolves around the concept of continuity of functions on composed intervals, specifically examining whether a function can be continuous on a union of disjoint intervals. The example used is the function f(x) = 1/x and its continuity on the intervals (-∞, 0) and (0, ∞).

Discussion Character

  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant questions whether f(x) = 1/x is continuous on the composed interval (-∞, 0) ∪ (0, ∞) or only on the individual intervals (-∞, 0) and (0, ∞).
  • Another participant references the general definition of continuity in topological spaces, suggesting that continuity does not require the domain to be connected, implying that the function can be continuous on the composed interval.
  • A further response reiterates that continuity is a local concept, stating that since f(x) is continuous at every point in ℝ - {0}, it is continuous on any subset of this domain, including the composed interval.
  • A later reply expresses appreciation for the question and the answers provided, indicating engagement with the discussion.

Areas of Agreement / Disagreement

Participants present differing views on the interpretation of continuity on composed intervals, with some asserting that continuity can be established over disjoint intervals while others focus on the local nature of continuity. The discussion remains unresolved regarding the implications of continuity in this context.

Contextual Notes

There are assumptions regarding the definitions of continuity and the nature of the intervals discussed. The implications of continuity in topological spaces and the local versus global nature of continuity are not fully explored.

ronaldor9
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Can a function be continuous on a composed interval? For example, if [tex]f(x)=\frac{1}{x}[/tex] then on the interval [tex](-\infty,0) \cup (0,\infty), f(x)[/tex] is continous? Or is the function [tex]f(x)[/tex] continuous on [tex](-\infty,0)[/tex] by itself and [tex](0,\infty)[/tex] by itself (If you don't get what I'm trying to say reply back)?
 
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If you study the general definition of continuity, using topological spaces, there is no requirement that the domain be a connected set. So in general, the answer to your question is yes.
 


ronaldor9 said:
For example, if [tex]f(x)=\frac{1}{x}[/tex] then on the interval [tex](-\infty,0) \cup (0,\infty), f(x)[/tex] is continous? Or is the function [tex]f(x)[/tex] continuous on [tex](-\infty,0)[/tex] by itself and [tex](0,\infty)[/tex] by itself?
It's the same. Continuity is a local concept: a function is continuous on some domain D if it is continuous at every point in D. Since your f is continuous at every point in [tex]\mathbb{R}-\{0\}[/tex], it is continuous on every subset [tex]D\subseteq\mathbb{R}-\{0\}[/tex], in particular on [tex]D= (-\infty,0) \cup (0,\infty)[/tex].
 


Great question with great answers!
 

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