How to do the following Lemma

[Cratylus]

This is not homework. I looked everywhere how to do it
Let {[imath]x_n:n\in Z[/imath] } be a sequence in A in R which is monotone increasing or monotone descreasing.
Let f { [imath]{x_n:n\in Z}[/imath] } by f([imath]x_n[/imath] )=n for each n. Then f is continuous on { [imath]x_n:n\in Z+[/imath] } and if this set is a closed subset of A f can be extended to a continuous function F:A->R

Hint: suppose our sequence is monotone increasing .The intervals [imath](x_n,x_{n+1})[/imath] and (n,n+1)are very much alike.. Patch together some restrictions of homeomorphism to get an extension of f

Sourace:A first source in topology .Robert Conover pg 144

 

[Deleted member 4993]

This is not homework. I looked everywhere how to do it
Let {[imath]x_n:n\in Z[/imath] } be a sequence in A in R which is monotone increasing or monotone descreasing.
Let f { [imath]{x_n:n\in Z}[/imath] } by f([imath]x_n[/imath] )=n for each n. Then f is continuous on { [imath]x_n:n\in Z+[/imath] } and if this set is a closed subset of A f can be extended to a continuous function F:A->R

Hint: suppose our sequence is monotone increasing .The intervals [imath](x_n,x_{n+1})[/imath] and (n,n+1)are very much alike.. Patch together some restrictions of homeomorphism to get an extension of f

Sourace:A first source in topology .Robert Conover pg 144

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING​

Please share your work/thoughts about this problem.

 

[Cratylus]

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING​

Please share your work/thoughts about this problem.

I was trying to post it but I ran out of time

My try:
Suppose we have an increasing sequence. Then if x_n < y_n implies f(x_n)< f(y_n) So by Def. of
continuous function if for each open set U of
Y,f^-1(U) is open in X
So f^-1(x_n) < f^-1(y_n)
implies f^-1(f(x_n)) < f^-1f((y_n))