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Let $x \in \mathbb{R}$. The finite complement topology on $\mathbb{R}$ is a collection of subsets of $U$ of $\mathbb{R}$ such that $U^{c}$ is finite or all of $\mathbb{R}$. Since $\{x\}^{c}$ is infinite in $\mathbb{R}$, $\{x\}$ is not open (not necessarily closed). This question arises as we were given the task of giving an example of a topological space that is not Hausdorff but does satisfy the $T_{1}$ axiom, that finite point sets are closed. The finite complement is not Hausdorff as given any $x_{1},x_{2} \in \mathbb{R}$, the only neighborhood in the finite complement topology which contains either point is all of $\mathbb{R}$. Thus, there does not exists neighborhoods $U_{1},U_{2}$ of $x_{1},x_{2}$ that are disjoint. So finally I come back to my question, is $\{x\}$ closed in the finite complement topology? My proof began claiming that it is not open (but not necessarily closed), however the solution I am looking at for this problem claims that the finite complement topology is not Hausdorff, for which I agree, and also satisfies the $T_{1}$ axiom, hence $\{x\}$ closed, which I have yet to prove. Since I am new to this, I am making an educated guess that there must be another way to prove that $\{x\}$ is closed or I am misinterpreting the meaning of open/closed sets in the finite complement topology. I would appreciate if someone would clear this up for me. Thank you in advance.

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    $\begingroup$ Yes, in finite complement topology the closed sets are exactly finite sets and whole space. $\endgroup$ – Wojowu Jan 28 '16 at 8:49
  • $\begingroup$ I suspect that you want the last word of the title to be closed, not open. $\endgroup$ – Brian M. Scott Jan 28 '16 at 8:52
  • $\begingroup$ Also remember that open is not the negation of closed, in this context. Sets can be both and sets can be neither. $\endgroup$ – Justpassingby Jan 28 '16 at 8:54
  • $\begingroup$ Yes, I just realized the mistake in the question and the mistake of thinking that open was the negation of closed. I have edited my question now so I think it makes more sense. $\endgroup$ – Jack Jan 28 '16 at 8:55
  • $\begingroup$ The complement of $\{x\}^c$ is $\{x\}$, which is finite. Therefore $\{x\}^c$ is open, i.e. $\{x\}$ is closed. $\endgroup$ – D_S Jan 28 '16 at 22:04
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The finite complement topology on any set is $T_1$. Let $x$ and $y$ be distinct points. Then $X\setminus\{y\}$ is an open nbhd of $x$ that does not contain $y$, and $X\setminus\{x\}$ is an open nbhd of $y$ that does not contain $x$. Here I’ve used what I consider the most usual definition of the $T_1$ property. If your definition of the $T_1$ property is that singleton sets are closed, you can adapt the argument very easily: for any $x\in X$, $X\setminus\{x\}$ is open, so its complement, $\{x\}$, is closed.

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  • $\begingroup$ Ah I see. So for $U$ to be a neighborhood in the finite complement topology, $U^{c}$ must be finite, i.e., finite point sets. Thus, since $U$ is open, $U^{c}$ (finite point sets), are closed. Is this right? $\endgroup$ – Jack Jan 28 '16 at 9:07
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    $\begingroup$ @Tim: $X\setminus U$ must be finite, but that doesn’t mean that it has to be a singleton: all singletons are finite, but so are all sets with two elements, three elements, etc. Apart from that, though, you’re okay: the complement of an open set is by definition closed, and in the finite complement topology every finite set is the complement of an open set, so every finite set is closed. $\endgroup$ – Brian M. Scott Jan 28 '16 at 9:09
  • $\begingroup$ Ok I edited my comment above to reflect your comment. Is it right now? $\endgroup$ – Jack Jan 28 '16 at 9:12
  • $\begingroup$ @Tim: Yes, it’s fine now. $\endgroup$ – Brian M. Scott Jan 28 '16 at 9:13
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Here is a better answer I have come up with:

Consider the finite complement topology on $\mathbb{R}$ and $x \in \mathbb{R}$. For any $x \in \mathbb{R}$ we have that ${\mathbb{R} - (\mathbb{R} - \{x\}) = \{x\}}$, so $(\{x\}^{c})^{c}$ is finite. Thus $\{x\}^{c}$ is open in the finite complement topology, hence $\{x\}$ is closed.

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Let ($\mathbb R, \frak T_f$) be a topological space,where $\frak F_f$ is a finite complement topology(or,Cofinite toplogy).

then,$\frak F_f$={ $A \subset \mathbb R:A^{c} $ is finite,or $A^{c}= \mathbb R$}.

Let $x \in \mathbb R$,now our aim is to show that $U^{c}$={$x$} is closed in ($\mathbb R, \frak T_f$) i.e., to prove $U$ is open in ($\mathbb R, \frak T_f$) .

Now,$U^{c}$={$x$},which is finite $\implies U \in \frak F_f \implies $ $U$ is open in ($\mathbb R, \frak T_f$) $\implies U^{c}$={$x$} is closed in ($\mathbb R, \frak T_f$).

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