First of all I apologise if this question will sound stupid. I'm approaching to the study of topological spaces and real analysis in a deeper way, that is trying to fill holes and doubts and this one has came in my mind.
I'll start with this: consider the set $A = \{\frac{1}{n+1}: n \in \mathbb{N}\} \subset \mathbb{R}$. In my definition, $0 \in \mathbb{N}$.Now $A$ is a set which I can write as a union of singlets, and each singlet is a closed set. So far so good, I know why they are closed and so on.
Now consider the function $f: [10, 11] \cup \{14\} \to \mathbb{R}$ with $f(x) = x^2$. The domain of $f$ is closed, since it's a finite union between a compact set and a singlet. Though, I just found out, reading over here and else, that actually $\{14\}$ is not a closed set: it's open!!
This is due to the topological definition of continuity I have read, that is: $f$ is continuous if the pre image of any open set of $\mathbb{R}$ is open.For my case let's take $(196-\delta, 196+\delta)$ where for example $\delta = 1/3$. This is an open set in $\mathbb{R}$, but the preimage which is $f^{-1}(196-\delta, 196+\delta) = 14$ that is $\{14\}$ which is apparently not open, for what I know of singlets. But this would mean $f$ is not continuous at $x = 14$.
Can someone please help me in understanding where I am wrong? I may suppose that the reason of what I did not understand may lie into "what topology" I am using, but here other doubts arise.
Taking back the set $A$, this is a subset of $\mathbb{R}$, so what topology am I using? The discrete one? After all every singlet of $A$ is closed.
But again, when studying the continuity of $f$ at $x = 14$, what topology am I using? Would the result change? I don't even know if it makes sense to use the discrete topology here...
What if I had $f: [10, 11] \cup \{14\} \to \mathbb{Q}$, how would things change?
Thank you for your time...