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TopNewestReframe and elaborate your question. – Snehal Shekatkar · 3 years, 10 months ago

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In my opinion, it's vacuously true, but I'm not sure. – Ivan Koswara · 3 years, 10 months ago

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– Mark Hennings · 3 years, 9 months ago

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.Log in to reply

– Kulkul Chatterjee · 3 years, 9 months ago

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?Log in to reply

– Snehal Shekatkar · 3 years, 9 months ago

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.Log in to reply

– Snehal Shekatkar · 3 years, 10 months ago

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-continuousLog in to reply