Proove each of the following:
(1) For any point p an element of R and N subset of R, N is a neighborhood for p in R* iff p is an element of N.
R represents the real numbers
R*={A is a subset of X|a is an element of A implies {a} is a subset of A} (discrete topology)
(2) For any point p an element of R, N is a neighborhood for p in R_* iff N=R
R_*={A is a subset of X|a is an element of A implies X is a subset of A} (indiscrete topology)
(3) For any point p an element of R, N is a neighborhood for p in R_fc iff p is an element of N and there is a countable set A a subset of R st N = (R\A).
R_fc={A is a subset of X|a is an element of A implies (X\A) is finite} (finite-complement topology)
(4) For any point p an element of R and any e > 0, if N_e(p) = (p – e, p + e), then N_e(p) is a neighborhood for p in (R,T_R)
T_R = {A subset of R|a is an element of A implies (a – e,a + e) is a subset of A for some e >0)
(5) For any point p an element of R and any e > 0, if N_e(p) = [p, p + e), then N_e(p) is a neighborhood for p in R_S
R_S = {A is a subset of R | a is an element of A implies [a, a + e) is a subset of A for some e > 0} (Sorgenfrey topology)
(6) For any point p an element of R and any e > 0, if N_e(p) = (p – e, inf), then N_e(p) is a neighborhood for p in R_U
R_U = { A is a subset of R | a is an element of A implies (a - e, +inf) is a subset of A for some e > 0} (upper topology)
Attempted proofs:
(1) Claim 1: If N is a nbhd of p in R*, then p is an element of N
Proof: Suppose N is a nbhd of p
We know G is an element of R* such that p is an element of G and G is a subset of N (definition of nbhd)
Hence p is an element of N
Claim 2: If p is an element of N, then N is a nbhd of p in R*
Proof: Suppose p is an element of N
Claim 2.1: There exists G as an element of R* such that p is an element of G and G is a subset of N
Proof: We know p is an element of N, N is a subset of N and N is an element of R*
Therefore, there exists G in R* such that p is an element of G and G is a subset of N
(2) Claim 1: If N is a nbhd of p in R_*, then N=R
Proof: Suppose N is a nbhd of p in R_*
Claim 1.1: N is a subset of R
(((((PROVE Claim 1.1))))))
Claim 1.2: R is a subset of N
(((((PROVE Claim 1.2))))))
Hence N =R
Claim 2: If N=R, then N is a nbhd of p in R_*
Proof: Suppose N=R
Claim 2.1: There exists G as an element of R such that p is an element of G and G is a subset of N
Claim 2.1.1: N is an element of R_*
(((((PROVE Claim 2.1.1))))))
We know p is an element of R
So p is an element of N and N is a subset of N
(3) Claim 1: If N is a nbhd of p in R_fc, then p is an element of N and there is a finite set A in R such that N=(R\A)
Proof: Assume N is a nbhd of p in R_fc
Claim 1.1: p is an element of N and there is a finite set A in R such that N=(R\A)
Proof:We know F is an element of R_fc such that p is an element of G and G is a subset of N (def of nbhd)
So p is an element of N
((((((Not sure about these lines))))
==>We know R_fc(A)={A is a subset of R|a is an element of A=>(R\A) is finite)}
==>Since p is an element of N, we know N is a subset of R and p is an element of N, so N=(R\A)
Claim 2: If p is an element of N and there is a finite set A in R such that N=(R\A), then N is a nbhd
(((((PROVE Claim 2))))))
(4) Claim 1: (a,b) is an element of T_R for all a,b which are elements of R
Proof:
Claim 1.1: (a,b) is a subset of R
Proof: have show in class
Claim 1.2: If p is an element of (a,b), then there exists an e>0 st (p – e, p+e) is a subset of (a,b)
Proof: Suppose p is an element of (a,b)
Take e=min{(p – a),(b – p)}
Claim 1.2.1: e>0 and (p – e, p+e) is a subset of (a,b)
We know a<p<b and e=p – a >0 or e=b – p >0
So e>0
e=p – a => or a=p – e => a is less than or equal to p – e
e=b – p >0 or p+e=b => b is greater than or equal to p + e
Clearly, p – e < p + e and (p – e, p+e) is a subset of (a,b)
I feel pretty good about (1) and (4) but wouldn't mind them double checked. I got stuck on the marked parts of (2) and (3) and could use some help on approaching (5) and (6)