Mathematics Asked by Christine on November 30, 2020
Reading the wikipedia page for "Topological Space", three axiomatizations are presented for defining topological spaces. They are presented as equivalent. The first is attributed to Hausdorff and uses neighbourhoods, the second, familiar from way back when, uses open sets, and the third uses closed sets.
Examples are given of topologies and non-topologies on the three point set, and these make perfect sense following the definition via open sets. In particular, {{1,2,3}, {1}} is a topological space, but {{1,2,3}, {1}, {2}} is not because {1,2}, the union of {1} and {2} is not in the space.
For the life of me, I can not see how the first example, {{1,2,3}, {1}}, fits the axiomatization by Hausdorff. Either, and most probably, I am not understanding that axiomatization, or that axiomatization is only for Hausdorff topologies and is not equivalent to the definition via open sets.
Here is the given axiomatization from Hausdorff, with 1-3 made slightly more concise:
The set of subsets of X given by $mathbf{N}(x)$ is called a topological set if:
By this axiomatization, wouldn’t $mathbf{N}(x) = {{1,2,3},{1}}$ fail the second requirement? ${1}subset{1,2}subset{1,2,3}$ and ${1}inmathbf{N}(x)$, but ${1,2}notinmathbf{N}(x)$.
Full disclosure: While rationally I fully expect to be told what it is I haven’t understood, in my heart I hope I have understood correctly and that the article is wrong-or-misleading because Hausdorff’s axiomatization is not equivalent.
Your example isn't quite right: ${{1},{1,2,3}}$ is not in fact a topology. You need to throw in the emptyset: ${emptyset, {1},{1,2,3}}$ is a topology.
The definition on wikipedia is correct - the issue is what exactly is meant when we say "$tau$ is a topology on $X$ in the sense of Hausdorff."
I suspect that you are getting tripped up by interpreting this as "$tau=ran({bf N})$ for some ${bf N}$ satisfying Hausdorff's axioms." But that's not correct. Instead, we need to introduce yet another notion:
$Usubseteq X$ is open according to ${bf N}$ iff for each $xin U$ we have some $Vin {bf N}(x)$ such that $Vsubseteq U$.
(On the wikipedia page, this is presented in the last paragraph of the "definition by neighborhoods" section.)
This notion is what provides the bridge between the "neighborhood function" idea of topology and the "set of open sets" idea of topology: when we say
"$tau$ is a topology on $X$ in the sense of Hausdorff,"
what we mean is
"there is some ${bf N}$ satisfying Hausdorff's axioms such that $tau$ is the set of open sets in the sense of ${bf N}$."
Note that if $U$ is open in the sense of ${bf N}$ and $xin U$ then $Uin {bf N}(x)$, but the converse is false: you want to interpret "neighborhood of $x$" as "set containing some open set containing $x$." It is true (as you observe) that ${emptyset,{1},{1,2,3}}$ is not $ran({bf N})$ for any ${bf N}$ satisfying Hausdorff's axioms, but that's irrelevant.
In particular, to show that $${emptyset, {1},{1,2,3}}$$ is a topology on ${1,2,3}$ in the sense of Hausdorff, we need to exhibit an $${bf N}:{1,2,3}rightarrowmathcal{P}({1,2,3})$$ satisfying Hausdorff's neighborhood function axioms such that the ${bf N}$-open sets are exactly $emptyset$, ${1}$, and ${1,2,3}$. This might look daunting, but luckily there's a general recipe for doing this. Given any collection $mathcal{S}$ of subsets of $X$, consider the function $${bf N}_mathcal{S}: xmapsto{Uinmathcal{S}: xin U}.$$ It turns out (and this is a good exercise) that this is the only thing we need to consider: $mathcal{S}$ is a topology on $X$ in the usual sense iff ${bf N}_mathcal{S}$ satisfies Hausdorff's neighborhood function axioms (in which case $mathcal{S}$ will be exactly the collection of open sets in the sense of ${bf N}_mathcal{S}$).
In this case we're led to consider the function ${bf N}: {1,2,3}rightarrowmathcal{P}({1,2,3})$ given by:
$1mapsto {{1},{1,2}, {1,3}, {1,2,3}}$,
$2mapsto{{1,2,3}}$,
$3mapsto {{1,2,3}}$.
It's easy to check that this ${bf N}$ satisfies Hausdorff's axioms and that the open sets in the sense of ${bf N}$ are exactly $emptyset$, ${1}$ and ${1,2,3}$
Correct answer by Noah Schweber on November 30, 2020
You misunderstood the definition: $N$ is a function of $X = {1,2,3}$ into the set of filters on $X$ such that for every $xin X$ $N(x)$ satisfies properties 1 and 4. The function maps a point $xin X$ to its neighbourhood filter.
If the open sets of $X$ are precisely $emptyset, X, {1}$, the function is given by $N(1) = {{1},{1,2},{1,3},{1,2,3}}$, $N(2)=N(3) = {{1,2,3}}$.
Answered by cuperius on November 30, 2020
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