I found this problem in chapter 7 of biglist. This is supposed to be an easy problem, but I have no idea how to approach it. The problem goes as follows:
Let $(X, \mathcal{T})$ be a topological space and let $A \subseteq X$. Let $i: A \to X$ be the inclusion function defined by $i(x) = x$. Show that the topology $A$ inherits from $X$ is the coarsest topology on $A$ such that $i$ is continuous.
Specifically, I want to know, the precise definition of a topology being coarse and how to start proving that some topology is coarser than others. I know about topologies refining each other. Is that what I am supposed to prove here? Any help is appreciated.
EDIT: (An attempt at a proof)
Let's first show that $i$ is continuous. Let $U$ be an open set in $X$. Then, $i^{-1}(U) = U \cap A$. Since $U$ is open in $X$ and $A \subseteq X$, the intersection of $U$ and $A$ is open in $X$. Therefore, $i^{-1}(U)$ is open in $A$, which shows that $i$ is continuous. Now we need to show that no strictly finer topology on $A$ makes $i$ continuous. Or in other words, we want to show if $\mathcal{T}_c$ is a topology on $A$ that makes $i$ continuous, then $\mathcal{T}_c \subseteq \mathcal{T}$, where $\mathcal{T}$ is the topology on $A$ inherited from $X$. Let $U \in \mathcal{T}_c$. Since, $i$ is continuous, $i^{-1}(U)$ must be open in $A$. But $i^{-1}(U) = U \cap A$. So, $U \cap A$ is open in $X$ with respect to the topology inherited from $X$. So, $U \cap A \in \mathcal{T}$. Thus $\mathcal{T}_c \subseteq \mathcal{T}$.
Feel free to comment :)
Let $\mathcal{T}_s$ denote the (subspace) topology that $A \subseteq X$ inherits from $X$. Let $\mathcal{T}_A$ be any topology on $A$ with respect to which the function $i : A \mapsto X$, defined by $i (a) = a$, is continuous.
Let $V \in \mathcal{T}_s$ be arbitrary. By definition, there exists $U \in \mathcal{T}$ such that $V = A \cap U$. We note that $i^{-1} (U) = A \cap i^{-1} (U) = i^{-1} (A \cap U) = A \cap U = V$ is contained in $\mathcal{T}_A$ since $i$ is continuous with respect to the topology $(A, \mathcal{T}_A)$. Thus, $\mathcal{T}_s \subseteq \mathcal{T}_A$. Since $\mathcal{T}_A$ was arbitrary, $\mathcal{T}_s$ is the coarsest topology on $A$ with respect to which $i$ is continuous. In other words, $\mathcal{T}_s$ is contained in every topology with respect to which $i$ is continuous.