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I have encountered in my introductory topology class an exercise problem which seems to contradict itself. We are told that a subset $S$ of a topological space $(X,\tau)$ is saturated if it is an intersection of open sets in $(X,\tau)$.

In part one we have to verify that every open set is indeed saturated. This is easily done - just choose an arbitrary open set from our topology and realize that $A \cap X=A$. Since $A$ and $X$ are both open sets of our topology, an arbitrary open set is a saturated one. So far so good.

In part two we ought to show that in a $T_1-$space every set is saturated. (a $T_1-$space is a one in which all of the singletons ${x}$ form a closed set. What's the point of this excercise? Since a $T_1-$space forms a topology, it immediately follows from part one that every set in our topology is saturated.

In part three however we need to present an example of a topological space which has at least one subset which is not saturated.

I suppose I am missing something, otherwise the questions wouldn't seem to strangely phrased to me.

Could you help me to decipher the real meaning of the exercise?

Thank you very much!

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    $\begingroup$ In part one you showed that all open sets are saturated. Now you need to show that all sets are saturated under an additional assumption. $\endgroup$ – Tobias Kildetoft Jul 4 '16 at 18:54
  • $\begingroup$ Also, for part one $A=A$ is really enough, since this is an intersection of one open set. If you want, $A = \bigcap_{i\in I} A_i$ with $I=\left\{1\right\}$, $A_1=A$. $\endgroup$ – Christoph Jul 4 '16 at 18:56
  • $\begingroup$ So by this one means every subset of $X$, not every subset of our topology? Because all of the subsets of our topology are opened... $\endgroup$ – whypi314 Jul 4 '16 at 18:56
  • $\begingroup$ Yes, every subset of a $T_1$-space is saturated! $\endgroup$ – Christoph Jul 4 '16 at 18:57
  • $\begingroup$ To be more precise: All elements of the topology are open. The elements of the topology are subsets of the space. $\endgroup$ – Tobias Kildetoft Jul 4 '16 at 18:59
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Part 1 Let $(X,\tau)$ be an arbitrarily topological space. If $U\in\tau$ is an open set, then it can be represented as the intersection of the one-element family $\{U\}$. Therefore, $U$ is saturated. (If you really want take the plural in “intersection of open sets” seriously, simply take $U=U\cap U$ or $U=U\cap X$.)


Part 2 If $(X,\tau)$ is $T_1$, then the singleton set $\{x\}$ is closed for any $x\in X$. Therefore, $\{x\}^{\mathsf c}$ is open. Now take any arbitrary subset $A\subseteq X$. If $A=X$, then $A$ is open, so that it is saturated by the previous part. If, on the other hand, $A\neq X$, then $A^{\mathsf c}$ is not empty. Now observe that $$A^{\mathsf c}=\bigcup_{x\in A^{\mathsf c}}\{x\},$$ whence it follows from De Morgan’s laws that $$A=(A^{\mathsf c})^{\mathsf c}=\left(\bigcup_{x\in A^{\mathsf c}}\{x\}\right)^{\mathsf c}=\bigcap_{x\in A^{\mathsf c}}\{x\}^{\mathsf c}.$$ Consequently, $A$ is an intersection of open sets and hence saturated.


Part 3 Let $X\equiv\{1,2\}$ and $\tau\equiv\{\varnothing, X\}$ be the indiscrete topology. Then, the only open sets are the empty set and $\{1,2\}$. Therefore, the singleton set $\{1\}$ can in no way be expressed as an intersection of open sets, so that $\{1\}$ is not saturated.


Extra The converse of Part 2 is true as well: If every subset of a topological space $(X,\tau)$, is saturated, then the space is $T_1$. To see this, consider any $x\in X$ and look at the set $\{x\}^{\mathsf c}$. If this set is to be saturated, then there must exist a family of open sets $\{U_i\}_{i\in I}$, indexed by some non-empty index set $I$, such that $$\{x\}^{\mathsf c}=\bigcap_{i\in I}U_i.$$ Now, each of the open sets $U_i$ contains $\{x\}^{\mathsf c}$. Therefore, $U_i$ equals either $\{x\}^{\mathsf c}=X\setminus\{x\}$ or the whole set $X$ for every $i\in I$. However, in order for the intersection to equal $\{x\}^{\mathsf c}$ as opposed to $X$, it must be the case that $U_i=\{x\}^{\mathsf c}$ for at least one $i\in I$. This implies that $\{x\}^{\mathsf c}$ is open, so that $\{x\}$ is closed. Consequently, $(X,\tau)$ is a $T_1$ space.

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  • $\begingroup$ Ok, many thanks - after I corrected my understanding of the question I arrived at the same conclusion. $\endgroup$ – whypi314 Jul 5 '16 at 6:58
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    $\begingroup$ Just one more question - does a saturated set imply a $T_1-$space? I have tried to come up with a simple counterexample, however, creating a saturated set with more than three points becomes pretty painful. The simplest cases seem to support the implication. $\endgroup$ – whypi314 Jul 5 '16 at 7:00
  • $\begingroup$ @whypi314 I added a proof of the converse statement. $\endgroup$ – triple_sec Jul 5 '16 at 16:17

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