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This question might be silly, however I'd like to be $100\%$ sure.

Suppose we have some topology $\tau$ on a set $X$ which gives us the $T_i$ space for some $i<6$. It's totally possible to improve topological properties (here I'd like to improve separation axioms) - simply consider basic anti-discrete topology, which consists only of $\emptyset$ and $X$.

My question is about the opposite situation. Is it possible to have topology $\tau_1$ on set $X$ and a bigger topology $\tau_2 \supset \tau_1$, such that $(X,\tau_1)$ is $T_i$ space while $(X,\tau_2)$ does not fulfill $T_i$ space axiom.

I'm pretty sure such situation is not possible, but if it is - please provide me with a proper counterexample.

For $i\leqslant 2$ I'm sure it's not the case, I wonder if making topology larger (which at the same time introduces new closed sets) can prevent $(X,\tau)$ from being $T_i$ space.

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  • $\begingroup$ Which cases do you speak about? If it's about $T_i$ for which $i<3$, then I've already pointed that out in my post. If you do speak about $i\geqslant 3$ please point out why it's so trivial, because I don't really see it. $\endgroup$ – I_Really_Want_To_Heal_Myself Jun 16 '17 at 20:05
  • $\begingroup$ did you read my answer? $\endgroup$ – Jorge Fernández Hidalgo Jun 16 '17 at 21:13
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    $\begingroup$ The point is, for $T_3$ and higher we have to consider closed sets as well, not just points. Adding open sets also adds closed sets, so more conditions to fulfill. $\endgroup$ – Henno Brandsma Jun 16 '17 at 22:06
  • $\begingroup$ It's a very good Q. $\endgroup$ – DanielWainfleet Jun 18 '17 at 2:39
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If a Hausdorf space is given a finer topology,
it's still Hausdorf. Proof is very simple.

The set of real numbers is regular and normal.
Add to the usual topolopy the set of rationals and
it is neither regular nor normal, yet still Hausdorf.

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Pick a set $X$ along with an equivalence relation $\sim$. Consider the topology $\tau$ on $X$ in which the open sets are the sets that are unions of equivalence classes.

In this space the closed and open sets coincide. This space is clearly $T_3$ .

Consider the refinement on $X$ which consists of taking one of the equivalence classes $C$ and refining the topology on $C$ into a non-$T_3$ topology.

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  • $\begingroup$ I think this approach works for the other ones also (Except $T_0,T_1,T_2$ and $T_{2.5}$) $\endgroup$ – Jorge Fernández Hidalgo Jun 16 '17 at 20:14
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An example where $\tau_1$ is $T_6$ but $\tau_2$ is NOT $T_6.$

Let $I=[0,1].$ Let $\tau_1$ be the usual topology on $I^2.$ Let $<_L$ be the lexicographic order on $I^2.$ That is $(x,y)<_L(x',y')\iff (x<x'\lor [x=x'\land y<y']).$ The topology $\tau_2$ induced on $I^2$ by $<_L$ is stronger than $\tau_1.$ ( If $A,B$ are open real intervals then $(A\times B)\cap I^2$ is a union of $<_L$-open intervals.)

Then $\tau_1$ and $\tau_2$ are $T_5$ topologies. (For $\tau_2,$ all linear spaces are $T_5$).

But $\tau_1$ is $T_6$ and $\tau_2$ is NOT $T_6. $

(I). $\tau_1$ is $T_6$ because it is metrizable.

(II). Notation: ]a,b[ and [a,b[ are real intervals. (a,b) is an ordered pair.

Let $A=I\times \{0,1\}.$ Then $A$ is $\tau_2$-closed but $A$ is not a $G_{\delta}$ set in the $\tau_2$ topology.

Proof: If $A\subset U\in \tau$ and $x\in [0,1[$ there is a $<_T$-open interval $J$ with $(x,1)\in J\subset U.$ And since $(x,1)$ is in the $\tau_2$-closure of $\{(x',y')\in I^2: (x,1)<_L (x',y')\},$ there is $(x',y')\in J$ with $x'>x.$ Since $J$ is a $<_T-$interval, therefore $]x,x'[\times I\subset J\subset U.$

So let $B(U)=\cup \{]x,x'[ \;: \;0\leq x<1\;\land \;]x,x'[\times I\subset U\}.$ With respect to the usual topology on $I,$ the set $B(U)$ is open and dense in $I.$ (If $0\leq x<y\leq 1$ there is $x'\in ]x,y[\cap B(U)$.)

So if $A\subset U_n\in \tau_2$ for $n\in \mathbb N$ then $$\cap_{n\in \mathbb N}U_n\supset (\cap_{n\in \mathbb N}B(U_n))\times I.$$ By the Baire Category Theorem applied to the usual topology on $I$ we have $ \cap_{n\in \mathbb N}B(U_n)\ne \phi.$

Therefore $\cap_{n\in \mathbb N}U_n\ne A.$

Remark: One-member subsets of $I^2$ are closed $G_{\delta}$ sets in $\tau_2$ but the closed set $A$ is not $G_{\delta}$ in $\tau_2$.

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  • $\begingroup$ I do like this one, although the notation makes it quite hard to see at the first glance. $\endgroup$ – I_Really_Want_To_Heal_Myself Jun 18 '17 at 21:31
  • $\begingroup$ I don't usually use $]a,b[$ for open interval but I also have ordered pairs involved. And I made it very detailed to make sure I got it right. The lex-order on $I^2$ is suggested as a hint in General Topology by Engelking in an exercise asking for a space with the property in my "Remark". $\endgroup$ – DanielWainfleet Jun 19 '17 at 16:14
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A $T_6$ topology $\tau$ and a finer topology $\tau'$ that is not even $T_4:$

Let $\tau$ be the usual (standard) topology on $\mathbb R^2.$

Let $S$ be the Sorgenfrey line (a.k.a. the lower limit topology on the set $\mathbb R$), which has a base of all real intervals $[x,y).$ Let $\tau'$ be the product topology on $S^2.$

A Q that has appeared repeatedly on this site is to prove that $S^2$ is not $T_4.$ The solution is to apply Jones' Lemma: If $X$ is a separable space with a closed discrete subspace $Y$ where the cardinal of $Y$ is $2^{\aleph_0}$ then $X$ is not normal. For the case $X=S^2$ let $Y=\{(x,-x): x\in \mathbb R\}.$

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