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X= {All continuous functions f:[0,1] $\rightarrow$ [0,1] } with the metric d(f,g)=$Sup_{x \in [0,1]}$|f(x)-g(x)|

A is a subset of the metric space (X, d) I'm trying to find the boundary of A which is a subset of the metric space.

A={f$\in$X : f(1/2)>1/2}}

Since g(x) isn't defined, should I just assume this is also on [0,1] $\rightarrow$ [0,1]?

So here f can be any function between from [0,1] $\rightarrow$ [0,1] just so long as you have (x>.5, y$\in$[0,1]).

Since X is the set of all functions, am I correct in thinking that the boundary points will be functions? For A would the answer look something like f(1/2)>1/2 or would I need to define a new function?

So my understanding is that I'm looking for functions such that dist(f(x), f(1/2)>1/2) = 0 = dist(f(x), f(1/2)$\le$1/2)

So would that mean the only boundary point is f(1/2)=1/2?

As far as I can see it's not possible to draw a diagram for this, is that correct? All I know is that the function can go through f(.5) until it's past x=.5.

I'm not looking for the answer but I'm not sure of what exactly I'm being asked to do or how the answer should look, so if someone could point me in the right direction I'd really appreciate it.

Thanks in advance!

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2 Answers 2

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Having a function $f$ on the boundary of $A$ means that if you take a ball around $f$ of any arbitrarily small radius $\epsilon > 0$, then the ball always contains a function in $A$ and another function not in $A$. In case you're not familiar with balls of continuous functions on a compact set using the sup metric, the ball of radius $\epsilon$ around $f$ is the set of all functions $g$ such that $|f(x) - g(x)| < \epsilon$ for all $x$. Work the cases $f(1/2) > 1/2$, $f(1/2) < 1/2$, and $f(1/2) = 1/2$ separately to determine which cases are on the boundary. As a further hint, you are right that it is $f(1/2) = 1/2$ that defines the functions on the boundary.

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what is your g(x)?

yes, the boundary of $A$ is again a subset of functions of $X$, and every boundary point is a function, too.

the notation $d(f(x),f(1/2)>1/2)$ does not make any sense! But your guess is correct: The boundary of $A$ contains all functions $f$ with $f(1/2)=1/2$.

To prove this, remember (one of) the definition(s) of a boundary point. A point (or in your case a function) $f$ is in the boundary of $A$ iff there are sequences $(f_n)\subset A$ and $(g_n)\subset X\backslash A$ such that $f_n\to f$ and $g_n\to f$. Here the convergence is meant with respect to $d$. So you should first answer the following questions:
- What does convergence in $d$ mean?
- What do you know about uniformly convergent sequences of continuous functions?
- What is the big advantage of uniform convergence?
- And what does uniform convergence have to do with your metric space?

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