Funtion continuity and open sets

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SUMMARY

The discussion centers on the relationship between continuity and the mapping of open sets in topology. It establishes that while a continuous function must map open sets to open sets, the converse is not true; a function can map open sets to open sets without being continuous. A constant function serves as an example of continuity without open set mapping, while a function defined on the interval [0, 1] with a discrete topology demonstrates open set mapping without continuity.

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  • Understanding of continuous functions in topology
  • Familiarity with open sets and their properties
  • Knowledge of discrete topology concepts
  • Basic proficiency in mathematical proofs and counterexamples
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Mathematicians, students of topology, and anyone interested in the foundational concepts of continuity and open sets in mathematical analysis.

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Homework Statement



Suppose that [tex]f : (X,d_X) \to (Y,d_Y)[/tex]. If [tex]f[/tex] is continuous,
must it map open sets to open sets? If [tex]f[/tex] does map open sets to
open sets must [tex]f[/tex] be continuous?

Homework Equations





The Attempt at a Solution



The answer to the first question is yes. The answer to the second question I guess is "no". Is this correct? How can I prove it?
 
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Neither is true. For example, a constant function is trivially continuous but maps any set, including any open set, to a singleton set which is not, generally, open.

Conversely, if B has the discrete topology, so that all sets are open, and A is a [0, 1] with the usual topology, the function f(x)= a for all [itex]0\le x< 1/2[/itex] and f(x)= b for all [itex]1/2\le 0\le 1[/itex], where a and b are distinct points in B, is not continuous but maps open sets to open sets.
 
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