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is point is continuous function ??

Note by Suryansh Tiwari
4 years ago

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Reframe and elaborate your question.

Snehal Shekatkar - 4 years ago

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The question is whether a function with a single-element domain is continuous. For example, is any function \(f : \{0\} \to \mathbb{R}\) continuous?

In my opinion, it's vacuously true, but I'm not sure.

Ivan Koswara - 4 years ago

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Indeed. If the domain contains a single element, then the inverse image \(f^{-1}(U)\) of any set is either empty or the whole domain. Both of these sets are open in the domain. Thus the inverse image of every set is open, and hence the inverse image of every open set is open, which means that \(f\) is continuous. This result holds no matter what set the codomain is, and no matter what topology it is equipped with.

Mark Hennings - 4 years ago

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Hey Ivan, is the given function a function at all ? I'm asking bcoz the ordered pairs ( i.e the function values in the cartesian plane) is of the form (0,R) R=real no. . All the ordered pairs have the same 1st element that is 0. So am I confusing something here?

Kulkul Chatterjee - 4 years ago

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@Kulkul Chatterjee No no.. he is talking about case where you pick a particular real number. So you will have only one pair \((0,r)\) and then indeed it is a function.

Snehal Shekatkar - 4 years ago

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I found this discussion on internet. May be this is where you should discuss it. :) http://math.stackexchange.com/questions/455296/can-a-function-with-just-one-point-in-its-domain-be-continuous

Snehal Shekatkar - 4 years ago

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