$(\mathbb{R}, \mathcal{T}) \cong (\mathbb{R}, \mathcal{T}')$ but $\mathcal{T} \neq \mathcal{T}'$

Question

Is it possible to have two homeomorphic spaces $(\mathbb{R}, \mathcal{T}) \cong (\mathbb{R}, \mathcal{T}')$ but $\mathcal{T} \neq \mathcal{T}'$?

Answer 1

Declare one of the topological space $(\Bbb R,\mathcal{T})$ to be the usual Euclidean one.

Choose any two distinct points $a,b\in\Bbb R$, and define another topological space by swapping their roles. Formally, consider the bijective function $f:\Bbb R\to\Bbb R$ with $f(a)=b$, $f(b)=a$ and $f(x)=x$ for all points $x\not=a$ or $b$. Define a topology $\mathcal{T}'$ on $\Bbb R$ by $U\in\mathcal T'$ if and only if $f(U)\in\mathcal T$. By construction, $f$ is a homeomorphism between $(\Bbb R,\mathcal T)$ and $(\Bbb R,\mathcal T')$.

But some connected open neighbourhoods of $a$ in $\mathcal T'$ would be of the form $\{a\}\cup(b-\delta,b)\cup(b,b+\varepsilon)$ for some small positive numbers $\delta,\varepsilon$, and such set is not in $\mathcal T$ usually. So $\mathcal T\not=\mathcal T'$.

Answer 2

There is a bijection $f:\mathbb R \to \mathbb R$ such that $f(0,1)=[0,1]$. Let $\mathcal T$ be the usual topology and $\mathcal T'$ consist of the sets $f(U)$ where $U \in \mathcal T$. Then $f$ is a homeomorphism from $(\mathbb R,\mathcal T) \to (\mathbb R,\mathcal T')$. However, $[0,1]$ is open in the second topology, not in the first.

Answer 3

The upper limit topology $\mathcal{T}_u$ and the lower limit topology $\mathcal{T}_l$ on $\mathbb{R}$ are distinct (see here), but homeomorphic via the homeomorphism $\varphi:(\mathbb{R},\mathcal{T}_u)\to(\mathbb{R},\mathcal{T}_l)$ sending $x\in\mathbb{R}$ to $-x$.

Answer 4

Think about the topology generated by a single point $x$: $\mathcal T_x=\{U\mid x\in U\text{ or } U=\varnothing\}$.

Now take $x\neq y$ and consider the two topologies concentrated on them.