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I have just started on learning topology. And I saw this question in Amann and Escher's Analysis I (Exercise 10, page 247):
Let X:={1,2,3,4,5} and
$\mathcal{T}:=\{\emptyset, X, \{1\}, \{3, 4\}, \{1, 3, 4\}, \{2,3,4,5\}\}$
Determine the closure of {2, 4, 5}
Since my understanding on the basic concepts are still quite weak, I'd like to ask whether my following reasoning steps are correct or not. Thank you!

My Reasoning:
All the sets in T are open sets.
Also, since $\{2, 3, 4, 5\}$ and $\{1\}$ are complement to each other and they are both elements in the topology. They are clopen sets.
So the closed sets with regard to this topology are
$\emptyset, X, \{2,3,4,5\}, \{1\}, \{1, 2, 5\}, \{2, 5\}$
Thus the closure of $\{2, 4, 5\}$ is the smallest closed set and at the same time its superset, that is $\{2, 3, 4, 5\}$

outis
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    That seems correct. – Giorgos Giapitzakis Jul 10 '21 at 21:54
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    @GiorgosGiapitzakis Thank you! May I also ask another question, that is, can I say in this case, the element 3 and 4 have the same neighbourhoods {3, 4}, {1, 3, 4}, and {2, 3, 4, 5}? – outis Jul 10 '21 at 21:57
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    That's correct, but you wrote too much. The fact that the sets ${1}$ and ${2,3,4,5}$ are clopen is irrelevant. – José Carlos Santos Jul 10 '21 at 21:58
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    The neighbourhoods you listed are open neighbourhoods of $3$ and $4$ and you also forgot $X$. – Giorgos Giapitzakis Jul 10 '21 at 22:00
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    @JoséCarlosSantos Thank you! For that, I just want to confirm my understanding of the concept open and closed set are correct since these concepts are still quite unfamiliar to me. – outis Jul 10 '21 at 22:01
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    @GiorgosGiapitzakis That's true! Thank you again! – outis Jul 10 '21 at 22:02
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    You wrote one thing that is correct but that suggests that there is some misunderstanding. You wrote that all the sets in $T$ are open sets, but it's more than that: the elements of $T$ are the open sets. – José Carlos Santos Jul 10 '21 at 22:07
  • @JoséCarlosSantos Yes, you are right! My formulation was wrong. Thank you for pointing that out! – outis Jul 10 '21 at 22:11

1 Answers1

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You're right about the closure and it's a valid reasoning (easy here because we have a full list of all closed sets). You could also say that $1$ has a neighbourhood $\{1\}$ that misses $\{2,4,5\}$ and so is not in the closure, and every neighbourhood of $3$ also contains $4 \in \{2,4,5\}$ so that $3$ is a limit point of your set, and $1$ is not, and so the closure only adds $3$ to your set. It of course gives the same result..

Henno Brandsma
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