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The following space is a version of example 71 from "Counterexamples in Topology". Let $X = \mathbb R$ be the real line with the Euclidean topology $\tau$. We define a new topology $\tau*$ on $\mathbb R$ to be the topology generated from the following basic open sets.

  1. Each point of the form $k+2^{-n}$, where $k,n \in \mathbb N$, is an open basic set.
  2. Open intervals of the form $(a,b)$ where $a<b$ are basic open sets.

I Claim that $\mathcal B$ is a basis for a topology, which implies that $\tau*$ is a topology on $\mathbb R$

Proof:

  1. $\bigcup \mathcal B = X$. This is obvious since, $\tau*$ is an extention of $\tau$.
  2. If $U,V \in \mathcal B$, and $x \in U \cap V$, then:

    a. If $x = k+2^{-n}$ for some $k,n \in \mathbb N$, then $x$ itself is a basic open set.

    b. If $x$ is not of the form $k+2^{-n}$ for some $k,n \in \mathbb N$, then, there are two intervals, $(a_u,b_u) \subset U,(a_v,b_v) \subset V$ which contains $x$, and the intersection $(a_u,b_u) \cap (a_v,b_v)$ is an open basic neighborhood, which contains $X$.

What do you think? Is the new proof correct??

Thank you!

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  • $\begingroup$ Everything up to 3 looks correct. For 3, what do you mean by a "discrete union"? $\endgroup$ Apr 6, 2014 at 16:10
  • $\begingroup$ @Omnomnomnom I think the OP meant "disjoint", not "discrete". $\endgroup$
    – augurar
    Apr 6, 2014 at 22:53
  • $\begingroup$ Thank you all for your answers. I have changed the proof accordingly. what do you think? $\endgroup$
    – topsi
    Apr 7, 2014 at 9:30

1 Answer 1

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The second part of your proof is not quite correct, since you have neglected to mention unbounded intervals. However, it is not necessary to attempt to characterize all open sets of the topology. The claim that unions of open sets are open follows trivially from the definition of $\tau^*$ as the collection of all unions of sets in the base.

For the third part of your proof, you need to show that the intersection of any two open sets in $\tau^*$ is open.

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  • $\begingroup$ Thank you. I have decided to proof it by "Base axioms". What do you think? Is the proof ok now? $\endgroup$
    – topsi
    Apr 7, 2014 at 9:30
  • $\begingroup$ @GeneralTopology 2b seems to skip a step - if $x$ is not of the form $k + 2^{-n}$, then $U$ and $V$ must both be open intervals. But if you are using this approach, you might actually want to offer a complete characterization of the intersection $U \cap V$, since there are only a small number of cases (empty, single point of the form $k + 2^{-n}$, or open interval). $\endgroup$
    – augurar
    Apr 8, 2014 at 21:39

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