MHB Discontinuous and continuous functions

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A function that is continuous at 0 but discontinuous at every other point can be defined as f(x) = x for rational x and f(x) = -x for irrational x. This function meets the criteria of being continuous at x=0 while failing to be continuous elsewhere. The discussion highlights confusion regarding the existence of limits at points other than 0. Participants suggest referring to the definitions of limits and continuity to clarify the concept. Understanding these definitions is crucial for grasping the behavior of such functions.
Carla1985
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I need to find a function that is continuous at 0 but discontinuous at every other point. IV been stuck on this for hours now :( thankyou
 
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Re: discontinuous and continuous functions

Carla1985 said:
I need to find a function that is continuous at 0 but discontinuous at every other point. IV been stuck on this for hours now :( thankyou

Hey Carla! ;)

How about:
$$f(x) = \left\{\begin{aligned}
x & \text{ if } x \in \mathbb Q \\
-x & \text{ if } x \in \mathbb R \backslash \mathbb Q
\end{aligned}\right.$$
 
Carla1985 said:
I need to find a function that is continuous at 0 but discontinuous at every other point. IV been stuck on this for hours now :( thankyou

Continuous at just one point ! , then how does the limit exist ?
 
Not a clue. I don't get it at all. The exact wording of a question, just in case iv got it wrong is: "give an example of a function defined on R which is continuous at x=0 and discontinuous at every other point of R". I like Serena, thank you :)
 
ZaidAlyafey said:
Continuous at just one point ! , then how does the limit exist ?

Consider the definition of the limit of a function, using $(\varepsilon, \delta)$-definitions.
Combine it with the definition of a continuous function in a point.
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

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