Question

Analysis help

Answer 1

@phi

Answer 2

There is no aditional info on X, maybe such that X is closed?

Answer 3

If that would be so, then you could use th fact that a continuous function is uniformly continuous on closed set by showing that the limit of any Couchy sequence belongs to X to prove the result

Answer 4

it means i will let {fn }be the cauchy sequence

Answer 5

so wat am i going to do with the sup{|f(x)-g(x)|}?

Answer 6

you are going to prove that under sup metric, convergence is uniform.
that is, if a sequence of continuous functions converges to a function, the convergence is uniform, and then we know that the uniform limit of continuous functions is a continuous function, therefore the metric is complete, as any cauchy sequence converges to something in \(C_d\)

Answer 7

let me give u my solution then check my solution

Answer 8

let {fn} be a cauchy sequence in X ,for every \[\epsilon >0\] let n>=N such that
\[|f _{n}(x)-g _{n}(x)|<epsilon\]

Answer 9

for any \[x \in X \] , we have
|\[f _{n}(x)-g _{n}(x)\]|\[\le ||f _{n}-g _{n}||_{\infty} < \in \]

i don't know wat to write next

Answer 10

@KingGeorge

Answer 11

the big confusion is that i have 2 different functions f and g that are uniform how to show both of them that they are uniform

Answer 12

Sorry, it's been a while since I've taken an appreciable amount of analysis, so I'm still trying to figure out all the definitions and how they relate to one another here.

Answer 13

It looks like you're almost there. You have to show that it's bounded and continuous to show that it's in \(C_b(X,R)\). That's almost the definition of continuous, and it should be fairly straightforward to show that it's bounded. I would start with continuous though, since it's a little more obvious from what you have.

Answer 14

@shevron