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Continuity in topology and the pasting lemma

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So right now I am reading on continuity in topology, which is stated as a function is continuous if the open subset of the image has an open subset in the inverse image..... that is not the issue.

I just read the pasting lemma which states:

Let X = A[itex]\cup[/itex]B, where A and B are closed in X. Let f: A-->Y and g: B-->Y. If f(x) = g(x) for every x [itex]\in[/itex] A[itex]\cap[/itex]B, then f and g combine to give a continuous function
h: X-->Y, defined by setting h(x)= f(x) if x[itex]\in[/itex]A and h(x) = g(x) if x[itex]\in[/itex]B.

Now they showed a couple of examples which I got except for this one:

l(x) = x-2, for x < 0 AND x+2 for x [itex]\geq[/itex]0

they go on to say that l(x) is not continuous because the inverse image of the open set (1,3) for instance is the nonopen set [0,1)

It's this last statment I am confused on how are they taking the inverse image of (1,3) and mapping it to the nonopen set? Seems like a simple enough concept but it is driving me bezerk.

Thanks in advance.
 

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  • #2
Office_Shredder
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Can you describe what you think the inverse image of (1,3) should be? You can calculate the inverse image of (1,3) under the function x+2 and intersect it with [0,infty), and then find the inverse image of (1,3) under the function x-2 and intersect that with (-infty,0), and then union those two things together.
 
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Can you describe what you think the inverse image of (1,3) should be? You can calculate the inverse image of (1,3) under the function x+2 and intersect it with [0,infty), and then find the inverse image of (1,3) under the function x-2 and intersect that with (-infty,0), and then union those two things together.


Ok let's see if this is the right thing to do:

for the function x+2 and the image (1,3)

I set 1 = x+2 and solved for x and got -1, I did the same for the other end of the interval: 3 = x + 2 and got 1, so I concluded that the interval is (-1,1), but my domain is specifically stated as
x [itex]\geq[/itex]0, therefore the smallest value this part of the function can take on is at 0. So that implies:

[0,1).

Now I did the same exact process for the x-2 portion and got an interval of (3,5) but that completely falls out of the specified domain of the function. Now you mentioned taking the intersection, would that entail taking the intersection of [0,1) [itex]\cap[/itex] (3,5) in which case it is empty so there is no values that can satisfy the pasting lemma. Hence l(x) is not continuous. Right?
 
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Ok let's see if this is the right thing to do:

for the function x+2 and the image (1,3)

I set 1 = x+2 and solved for x and got -1, I did the same for the other end of the interval: 3 = x + 2 and got 1, so I concluded that the interval is (-1,1), but my domain is specifically stated as
x [itex]\geq[/itex]0, therefore the smallest value this part of the function can take on is at 0. So that implies:

[0,1).
This part looks good
Now I did the same exact process for the x-2 portion and got an interval of (3,5) but that completely falls out of the specified domain of the function. Now you mentioned taking the intersection, would that entail taking the intersection of [0,1) [itex]\cap[/itex] (3,5) in which case it is empty so there is no values that can satisfy the pasting lemma. Hence l(x) is not continuous. Right?
This part less so. If x-2 lies in (1,3) then x is in (3,5). But you're only interested in x-2 when x<0, so there are NO values of x<0 for which f(x) lies in (1,3)

So the upshot is:
For x ≥ 0, if x is in [0,1) then f(x) is in (1,3).
For x < 0, there are no values of x such that f(x) is in (1,3).

How do you combine those two pieces of information?
 
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This part looks good


This part less so. If x-2 lies in (1,3) then x is in (3,5). But you're only interested in x-2 when x<0, so there are NO values of x<0 for which f(x) lies in (1,3)

So the upshot is:
For x ≥ 0, if x is in [0,1) then f(x) is in (1,3).
For x < 0, there are no values of x such that f(x) is in (1,3).

How do you combine those two pieces of information?
Well trying to go off of one of your previous suggestions, you mentioned I should union the sets together. So then would it be:

[0,1) [itex]\cup[/itex] (Empty Set) (can't find the empty set sign)

Now the reason I say empty set is because of the fact there are no values for x < 0

now the union of those two sets is [0,1) and from here you can see any value used in that interval doesn't provide equality between x-2 and x+2.
 
  • #6
Office_Shredder
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Well trying to go off of one of your previous suggestions, you mentioned I should union the sets together. So then would it be:

[0,1) [itex]\cup[/itex] (Empty Set) (can't find the empty set sign)

Now the reason I say empty set is because of the fact there are no values for x < 0
Do you understand why unioning them is the right thing to do?

now the union of those two sets is [0,1) and from here you can see any value used in that interval doesn't provide equality between x-2 and x+2.
This isn't a terribly relevant fact, all that matters is that you took the inverse image of an open set and you got [0,1), which is not open.
 
  • #7
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Do you understand why unioning them is the right thing to do?



This isn't a terribly relevant fact, all that matters is that you took the inverse image of an open set and you got [0,1), which is not open.
No I don't see why unioning them is the right thing to do, beyond perhaps trying to include everything in the set.....

Ok, so I took this image and got a set which was not open, but the only reason it was not open was because of the conditions on the function. If I didn't know the conditions of the function is there a way to deduce if a set is closed, open, etc?
 
  • #8
Office_Shredder
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No I don't see why unioning them is the right thing to do, beyond perhaps trying to include everything in the set.....

Ok, so I took this image and got a set which was not open, but the only reason it was not open was because of the conditions on the function. If I didn't know the conditions of the function is there a way to deduce if a set is closed, open, etc?
If you don't fully know the function I don't see how you could decide on what the inverse image is.

As to why you unioned them, you should go back to what the definition of an inverse image of a set is and try to work out why what we did makes sense as a procedure.
 
  • #9
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As to why you unioned them, you should go back to what the definition of an inverse image of a set is and try to work out why what we did makes sense as a procedure.
Well the definition of an inverse image is, the set U[itex]\subset[/itex]f-1(X) such that f(U) [itex]\subset[/itex] Y

and now the reason we union the sets is (to build on the vague explanation I gave) to include all of the elements that satisfy the condition of the inverse image. That is, for every element
"b" [itex]\in[/itex] f(x) [itex]\subset[/itex]Y, there exists an element
"a" [itex]\in[/itex] f-1 such that f(a) = b [itex]\in[/itex] f(x)
 

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