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My attempt to prove the statement of the title: if $X$ is infinite and $T_1$ with topology $\tau_1$ then any basis for $X$ is infinite too.

If $X$ is $T_1$ then for any $x, y\in X$ with $x\ne y$ exists some open sets $U$ and $V$ such that $x\in U \land y\notin U$ and $x\notin V \land y\in V$.

$\color{red}{(1)}$ If you extend this analysis for a finite subset $F\subset X$ you can create a collection $\mathcal{U}$ of open sets where $\forall x_i,x_j\in F,\ i\ne j,\ \exists U_i\in\mathcal{U}:\ (x_i\in U_i)\land (x_j\notin U_i)$. Then $F\bigcap(\bigcap_{i=0}^{n} U_i)=\varnothing$ and $\mathcal {U}=\{U_i\}$ but $\bigcap_{i=0}^{n} U_i\in\tau_1$ due the definition of topology (and remember $F$ is finite).

Then:

  1. If exists some countable infinite collection of disjoint open sets for the basis of $X$ then if we take some point belonging to any of these disjoint basic open sets then the resulting subspace is obviously homeomorphic to $\Bbb N$ with the discrete topology.

  2. If doesnt exist an infinite collection of disjoint basic open sets we have that exist an infinite collection of non-disjoint open sets. If we take infinite countable points each one belonging to a different but non-disjoint open set and cause the space is $T_1$ we have that for every open set at most a finite number of points ($F$ in the expression $\color{red}{(1)}$) not belong to them cause the statement on $\color{red}{(1)}$, so the subspace is homeomorphic to $\Bbb N$-cofinite.

Question:

  1. I feel my proof correct but not enough clear, maybe you can point how to clear it or if it lacks something?

  2. The expression on $\color{red}{(1)}$ is enough understandable?

Thank you in advance.


EDITION: the expression $\color{red}{(1)}$ is very hard to write it correctly to me by now, so Im going to describe in words: you can extend the definition of $T_1$-space not only to some $x,y\in X$ if not to $x_1,x_2,x_3,...,x_n\in X$. If you make all considerations then exist some $U_i$ for every $x_i$ where $x_i\in U_i$ but the others points does not belong to $U_i$. These collection of $U_i$ I call $\mathcal{U}$.

The intersection of every $U_i$ belongs to $\tau_1$ because is a finite intersection of open sets, and no one $x_i$ belongs to this intersection, obviously, and these $x_i$ are finite. This is what I wanted express on $\color{red}{(1)}$, sorry for the inconvenience :S

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  • $\begingroup$ I'm afraid I can't provide an answer to question 1, because the answer to question 2 is a very big NO. I can't even begin to figure out what in the world this is supposed to mean. I understand that you are not working in your native language here, so there is certainly some translation issues, but unfortunately, they are so great I can't figure it out. And what possible reason did you have for hiding part of it? $\endgroup$ Commented Dec 19, 2015 at 16:02
  • $\begingroup$ Oh, I just dont wanted to "hide" it @PaulSinclair, I just feel that it may be too much explanation. I will unhide it and try to change the notation, it is really wrong, yes. $\endgroup$
    – Masacroso
    Commented Dec 19, 2015 at 16:07
  • $\begingroup$ Let's start with $\{x_i\}$. You make use of it without properly defining what the notation means. It just suddenly shows up in your definition of $\mathcal U$ without explanation. $\endgroup$ Commented Dec 19, 2015 at 16:10
  • $\begingroup$ I edited my question @PaulSinclair, I hope this time it will be more clear, sorry man, I dont find the way to express my mind :S $\endgroup$
    – Masacroso
    Commented Dec 19, 2015 at 16:41
  • $\begingroup$ @Jake1234 - You are correct. My alternate approach was false. I thought I understood the cofinite topology, but apparently had not given it as much thought as I should have. I have deleted the false post. (Alas, if I had gulled one more person into upvoting it before you pointed out the mistake, I could have gotten another badge out of it...) $\endgroup$ Commented Dec 20, 2015 at 19:23

2 Answers 2

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Personally, I don't understand the $2.$ in your approach. The expression $\color{red}{(1)}$ is understandable, but the second part of your proof I do not understand.

An attempt, which is quite possibly wrong, so please take it with a grain of salt:

1st case - there exists a countable infinite subset $A$, such that it has cofinite topology, in that case we are done.

2nd case - such a subset does not exist, therefore in each infinite countable subset, there exists an element and open neightborhood of that element, such that it is not cofinite.

When we now take any countable infinite subset $A$, we can pick an element that has a non-cofinite open neightborhood, and we can call these $a_1$ and $U_1$. Consider the set $A \setminus U_1$ - it is still an infinite, countable subset, so we can again pick an element $a_2$ and an open neightborhood $U_2$, such that $(A \setminus U_1) \setminus U_2$ is infinite. By taking $U_2' = U_2 \cap (X \setminus \{a_1\} )$ ($(X \setminus \{a_1\} )$ is an open set in $T_1$) and $U_1' = U_1$ we have open neightborhoods of $a_1$ and $a_2$ that do not contain the other respective element.

We repeat this process, and define in this way a sequence $\{a_i: i \in \mathbb{N} \}$ of distinct points in $X$. For any element $a_i$ in this sequence, we have an open neightborhood $U_i'$, such that $U_i'$ contains no prior elements in the sequence (because there's finitely many of them, we can get rid from $U_i$ by using $T_1$ property of $X$), and all the further elements are defined to be from $A\setminus U_i$, therefore it doesn't contain any of them either. So the sequence has discrete topology with respect to $X$ - therefore, we have found an infinite countable subset with discrete topology, and so we are done.

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  • $\begingroup$ I will add some explanation to 2, and I will try to understand your proof. $\endgroup$
    – Masacroso
    Commented Dec 21, 2015 at 15:58
  • $\begingroup$ Your argument works (if we assume enough choice; if we reject choice with enough energy, there are infinite sets that don't have countably infinite subsets, then the assertion is false; and for the construction of the infinite discrete subset, we also need some choice [countable/dependent choice, ask Asaf if you want to know exactly how much choice is required]; but of course it's standard to assume full choice [everything else is counterintuitive and has weird consequences]). Perhaps the argument can be made clearer by assuming from the start that $X = \mathbb{N}$. $\endgroup$ Commented Dec 21, 2015 at 16:18
  • $\begingroup$ @DanielFischer Your comment is perhaps a bit pedantic. There is no claim that this is a Choiceless proof. Furthermore, if the exercise is to show that every infinite T₁ space has a subspace homeomoprhic to $\mathbb{N}$ with either the cofinite or discrete topology, then we have to assume that every infinite set has a countably infinite subset (i.e., there are no infinite Dedekind finite sets, which is somewhat weaker than the Axiom of Countable Choice). $\endgroup$
    – user642796
    Commented Dec 21, 2015 at 16:49
  • $\begingroup$ @Arthur Sure it's pedantic. But the pedantic part is just a parenthetical remark, safe to be ignored by everybody not interested in pedanticism ;) $\endgroup$ Commented Dec 21, 2015 at 16:53
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To address the questions you asked (sorry to be so late with this, but I've only had very limited time the last couple of days):

First, your expression $\color{red}{(1)}$ can be stated as:

Given a finite set $F\subseteq X$, for each $x \in F$, there is an open neighborhood $U_x$ of $x$ such that for all $y \in F, y \ne x \implies y \notin U_x$.

You can define $\mathcal U_F = \{U_x\ |\ x \in F\}$. I put the $F$ subscript on to remind you that $\mathcal U$ is defined from knowing $F$ first.

In point 2, you seem to be saying that in the case where $X$ does not contain an infinite collection of pairwise-disjoint empty sets, we can choose a countably infinite collection of open sets $\mathcal A$ and for each $A \in \mathcal A$, a point $x_A \in A$, such that if $A \ne B$, then $x_A \ne x_B$. However, like Jake1234, I don't see how $\color{red}{(1)}$ shows that $\{x_A\ |\ A \in \mathcal A\}$ has a cofinite subspace topology.

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  • $\begingroup$ Yes, thank you. I noticed the other day that what I did was wrong. I will try to improve my way to write and think these kind of topological problems. Thank you very much for your patience. $\endgroup$
    – Masacroso
    Commented Dec 22, 2015 at 12:15

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