Id $: \mathbb{R} \rightarrow \mathbb{ R}$ is the identity mapping then choose the correct statement

Question

Let $τ_1$ be the topology on $\mathbb R$ generated by the base $\mathcal B = \{[a, b):a,b\in\mathbb{R}\text{ and }a<b\}$.

Let $τ_0$ be the standard topology on $\mathbb{R}$ and let Id $: \mathbb{R} \rightarrow \mathbb{ R}$ be the identity mapping.

Choose the correct statement:

$(a)$ Id $: (\mathbb{R}, τ_1) \rightarrow (\mathbb R, τ_0)$ is continuous

$(b)$ Id $: (\mathbb{R}, τ_1) \rightarrow (\mathbb R, τ_0)$ is an open mapping

My attempt I thinks both option a) and b) are correct because lower limit topology is finer then usual topology.

Is it true ?

Any hints/solution will be appreciated.

Thank you.

Answer 1

If $U \in \tau_0$ then we can write $U=\cup (a_n,b_n)$ so $U=\cup_n \cup_j [a_n+\frac 1 j,b_n)$. Hence $U \in \tau_1$. This proves a). Since $[0,1) \in \tau_1$ but $[0,1) \notin \tau_0$ it follows that b) is false.

Answer 2

(a) is correct, but if a continuous bijective map is also open, then it is a homeomorphism. Do you think this is a homeomorphism?

Answer 3

In general:$$\mathsf{Id}:(X,\tau)\to(X,\rho)\text{ is open }\iff\mathsf{Id}:(X,\rho)\to(X,\tau)\text{ is continuous}\iff\tau\subseteq\rho$$ Draw conclusions.