Let $X$ be a compact space and $A \subset X$ be a closed subspace. Then we define $C_{\mathbb{R}}(A)$ to be the set of all continuous functions $f : A \to \mathbb{R}$.
Suppose we have a metric $d : C_{\mathbb{R}}(A) \times C_{\mathbb{R}}(A) \to \mathbb{R}$. Then I have been tasked with showing that all Cauchy-sequences in the metric space $(C_{\mathbb{R}}(A), d)$ converge.
I feel quite insecure in topology, but my idea was to show that if a Cauchy-sequence $(f_n)$ converges to a function $f$, then $f$ is also continuous. I would use the definition "$f:A \to \mathbb{R}$ is continuous on $A$ if for all open sets $O \in \mathbb{R}$, $f^{-1}(O)$ is open in $A$. This is where I need corrections, so please shoot down my claims if they are not true:
This requires for $A$ and $\mathbb{R}$ to have topologies. $A$ being compact, it already has a topology, call it $T$, and for $\mathbb{R}$ we may just use the order topology
Continuity of $f$ depends on the topologies defined on $A$ and $\mathbb{R}$. It is, however, sufficient to prove continuity of $f$ using just the topologies picked in $1)$, since then it would also hold for arbitrary topologies
Especially $2)$ is what I'm in doubt of. If it turns out to hold, then I would appreciate it if you would take the time to formulate an explanation as to why. Thank you very much. Regards, kasp9201.
The set-up is a bit vague, but I'll give it a shot:
Showing that all $d$-Cauchy sequences converge (under $d$ of course) entails that we start with a sequence of $(f_n)$ of functions from $C_{\mathbb{R}}(A)$, so all $f_n : A \to \mathbb{R}$ are continuous (in the subspace topology of $A$ w.r.t. $X$, and the reals in the standard/order topology), and from Cauchyness somehow (most of the time in such situations we have pointwise Cauchyness and we use completeness of the reals) produce an $f: A \to \mathbb{R}$ which must also be shown to be in $C_{\mathbb{R}}(A)$, so continuous too, and show that $d(f_n,f) \to 0$. There are no arbitary topologies here: $A$ has a given topology from being a subset of $X$ (which implicitly has a topology as you call $X$ compact), the reals always have their standard topology unless explicitly stated otherwise etc. Cauchy and convergence must be meant under $d$ (that's what completeness in a metric means).
So you must produce a candidate limit which you must prove to be in the space $C_{\mathbb{R}}(A)$, hence continuous.
We could help you better if things were more concrete and explicit.