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Math Help - Prove Closed Neighborhoods in R's Topologies

  1. #1
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    Prove Closed Neighborhoods in R's Topologies

    (1) For any point p, where p is an element of R, {p} is an open-and-closed neighborhood of p in R*
    R represents the real numbers
    R*={A is a subset of R|a is an element of A implies {a} is a subset of A} (discrete topology)
    (2) For any point p, where p is an element of R, N is a closed neighborhood of p in R_* iff N=R
    R_*={A is a subset of R|a is an element of A implies R is a subset of A} (indiscrete topology)
    (3) For any point p, where p is an element of R, and any e>0, [p-e, p+e] is a closed neighborhood of 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)
    (4) For any point p, where p is an element of R, and any e>0, [p, p+e] is an open-and-closed neighborhood of 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)
    (5) F is closed in R_fc iff F is finite or F=R
    R_fc={A is a subset of R|a is an element of A implies (R\A) is finite} (finite-complement topology)
    (6) N is a closed neighborhood of p in R_fc iff N=R
    (7) A nonemepty set F is closed in R_U iff F=(-infinity,b] for some b, which is an element of R, or F=R
    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)
    (8) N is a closed neighborhood of p in R_U iff N=R
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  2. #2
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    Re: Prove Closed Neighborhoods in R's Topologies

    Was this question to determine whether or not these are true? Most of them are not true.

    Quote Originally Posted by Kiefer View Post
    (1) For any point p, where p is an element of R, {p} is an open-and-closed neighborhood of p in R*
    R represents the real numbers
    R*={A is a subset of R|a is an element of A implies {a} is a subset of A} (discrete topology)
    So you are asked to show that every singleton set, {p} is both open and close in the discrete topology.
    By definition of "discrete topology", every singleton set is open. Because all unions of open sets are open, and every set is is a union of singleton sets, every set is open. Now, what is the definition of "closed set"?

    (2) For any point p, where p is an element of R, N is a closed neighborhood of p in R_* iff N=R
    Assuming you are still talking about the discrete topology, this is NOT true!

    R_*={A is a subset of R|a is an element of A implies R is a subset of A} (indiscrete topology)
    So this is a new topology. The condition that "a is an element of A implies R is a subset of A" (and, of course, "R is a subset of A" and "A is a subset of R" means that A= R) says that the only members of the "indscrete topology" are R itself and the empty set.

    (3) For any point p, where p is an element of R, and any e>0, [p-e, p+e] is a closed neighborhood of p in (R, T_R)
    Again, this is NOT true. A "closed neighborhood of p" is a closed set that contains p. And you just said that the only closed sets are R and the empty set.

    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)
    So if a is in an open set, there exist some (perhaps very small) interval, (a- e, a+e), that is a subset of A.

    (4) For any point p, where p is an element of R, and any e>0, [p, p+e] is an open-and-closed neighborhood of p in R_S
    Again, NOT true.

    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)
    (5) F is closed in R_fc iff F is finite or F=R
    R_fc={A is a subset of R|a is an element of A implies (R\A) is finite} (finite-complement topology)
    (6) N is a closed neighborhood of p in R_fc iff N=R
    (7) A nonemepty set F is closed in R_U iff F=(-infinity,b] for some b, which is an element of R, or F=R
    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)
    (8) N is a closed neighborhood of p in R_U iff N=R
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