Does introducing new open sets in topology improves it's "T"-axiom properties?

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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 $\tau_2$.

Answer 4

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 $Y=\{(x,-x): x\in \mathbb R\}.$