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I was looking at this question from 7 years ago and quite confused: Determine the closure of the set $K=\{\frac{1}{n}\mid n\in\mathbb N\}$ under each of topologies

The question is to find the closure of $K=\{\frac{1}{n}, n \in \mathbb{Z}^+\}$ in the five topologies

  • standard
  • $\mathbb{R}_k$
  • finite complement
  • upper limit topology
  • generated by $\beta = \{(-\infty, a)\}$

I was able to figure out $T_1, T_3,$ and $T_5$ just fine, but I am a little confused on $T_4$. I know $T_2 \subset T_1 \subset T_4$ and in the answer they say that since $T_2 \subset T_1, \bar{K}_{T_2} \subset \bar{K}_{T_1}$. Why does this argument not work for $T_4$ since $\{0\}$ is in $\bar{K}_{T_1}$?

Also, is the argument for $T_2$ basically that if you consider $0 \in \{(-1,1) - \{\frac{1}{k}\}_{k=1}^\infty \} = B,$ then $B \cap K = \phi$. Then none of the terms of the form $\{(a,b) - \{\frac{1}{k}\}_{k=1}^\infty \}$ conatin $K$ so you only have to consider the open intervals $(a,b)$ which is just the basis from the standard topology so the closure is a subset of $K \cup \{0\}$ so its just $K$?

John Smith
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1 Answers1

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T4. Because (-1,0] is an open nhood of 0 that misses K.

T2. Yes, there's an open nhood of 0 that misses K.

William Elliot
  • 17,598
  • Thanks, I get that much for $T_4$, but is that argument that the closure of a subset is a subset of the closure not always valid or was I wrong in finding that $T_1 \subset T_4?$ – John Smith Feb 25 '19 at 03:02
  • I do not understand that subset voodoo. Yes 1 is coarser than 4. – William Elliot Feb 25 '19 at 07:52