Question on topology. A function f :R-R is continuous if for each open set O in R the inverse image f^-1(O) is also an open set. I dont understand how this works, ok suppose I have a discontinuous function (removable discontinuity).

Question

Question on topology. A function f :R->R is continuous if for each open set O in R the inverse image f^-1(O) is also an open set. I dont understand how this works, ok suppose I have a discontinuous function (removable discontinuity).

anonymous · Answer

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anonymous · Answer

so f^-1( (1,3) ) = (1,2) U (2,3)

anonymous · Answer

i think you can  prove this by asuming the contrary

anonymous · Answer

man (lady) this is an excellent question!

anonymous · Answer

hello

anonymous · Answer

and the "answer" is that your "function", the one with a removable discontinuity, is not in fact a function on the interval in question because it is not defined at 2.

anonymous · Answer

yes but the preimage is open , so i dont see a problem

anonymous · Answer

no the function .... oh

anonymous · Answer

a better and fuller explanation of this exact question is here: http://thetwomeatmeal.wordpress.com/2010/11/04/continuous-maps-and-homeomorphisms/

anonymous · Answer

ok wait i will redefine it then

anonymous · Answer

it really is a great question.  if you redefine to be something else at 2, say 7, then you should see that it works.

anonymous · Answer

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