analysis001 said:Homework Statement
I'm trying to do a problem, and in order to do it I need to find a function f:R→R which is continuous on all of R, where A[itex]\subseteq[/itex]R is open but f(A) is not. Can anyone give an example of a function that satisfies these properties? I think once I have an example I'll understand things a lot better. Thanks.
analysis001 said:Oh ok would f(A)=sin(A) work?
Dick said:It would. Now you have to give an example of an open set A such that f(A) isn't open.
analysis001 said:If I take A to be the interval (-[itex]\pi[/itex],[itex]\pi[/itex]) which is open, then sin(A) would be on the interval [-1,1], which is closed. Am I understanding this correctly?
That's not true. ##\sin(\pi/2) = 1##, ##\sin(-\pi/2) = -1##Dick said:Almost, but sin((-pi,pi))=(-1,1) which is an open interval. The endpoints aren't included. Try sin((0,pi)).
Dick said:Almost, but sin((-pi,pi))=(-1,1) which is an open interval. The endpoints aren't included. Try sin((0,pi)).
analysis001 said:I don't get why sin((-[itex]\pi[/itex],[itex]\pi[/itex])=(-1,1). I thought because sin([itex]\pi[/itex]/2)=1 and since [itex]\pi[/itex]/2[itex]\in[/itex] (-[itex]\pi[/itex],[itex]\pi[/itex]) and also because sin(-[itex]\pi[/itex]/2)=-1 and since -[itex]\pi[/itex]/2[itex]\in[/itex] (-[itex]\pi[/itex],[itex]\pi[/itex]) then sin((-[itex]\pi[/itex],[itex]\pi[/itex])=[-1,1].
If sin((0,[itex]\pi[/itex]) then it equals (0,1] which is neither open nor closed. I'm not sure where my reasoning is wrong.
A continuous function is a type of mathematical function that has a smooth and unbroken graph. This means that as the input values of the function change, the output values change in a predictable way without any abrupt jumps or breaks. Mathematically, this can be represented as: limx→a f(x) = f(a), where a is any value within the domain of the function.
An open set is a set of points on a graph that does not include its boundary points. In other words, all the points within an open set are contained within the graph, but none of the points on the edge or boundary of the graph are included. This can be represented as x < a < y, where x and y are two points within the set and a is any value between them.
Continuous functions and open sets are closely related because a function is considered continuous if and only if the inverse image of any open set is an open set. This means that for any input value within an open set, the output value will also be within an open set. In other words, the function will not have any abrupt changes or breaks within the open set.
To determine if a function is continuous using open sets, we can use the definition of continuity which states that a function is continuous at a point if and only if the limit of the function at that point exists and is equal to the value of the function at that point. If we can show that the inverse image of any open set is an open set, then we can conclude that the function is continuous.
Continuous functions and open sets are used in various fields such as physics, engineering, and economics to model and analyze real-world phenomena. For example, in physics, continuous functions are used to describe the motion of objects and open sets are used to define the boundaries of a system. In economics, continuous functions and open sets are used to analyze market trends and make predictions about the behavior of consumers and businesses.