Thanks for your prompted help. Based on your definition of $d(x,C)$, $f(x)$ should be obviously continuous since the ratio on the RHS is just a well-defined number. However, I'm not sure how f(x) can be guaranteed to be $>=0$, since $x$ can be some points out of $C$ just like the given point $m$?

For part (b), we can choose a new function f different from what it was in part (a), as long as it satisfies the conditions in part (a).

Oh yes, you're right. the core idea works, but you need to adapt the function a little. Now, can you show me a function in $\mathbb{Q}$ at value in {0,1}, not constant, and continuous?

sorry but I can't:( Can you show such "weird" function?

The idea is to "cut" at an irrationnal. f(x) = 1 if x² > 2 and f(x) = 0 if x² < 2. But I need to think for the general case

btw, do you see why we need the condition $C$ is closed? I don't see its essence in the proof for part (a) yet.

If C is open and m is on the adherence of C, such function cannot exist.

why not though? It stills satisfies the continuity and the other 2 properties. The way d is defined works well for the caseyou mentioned...

No, it doesn't work, because $d(m,C) = 0$. If m is in the adherence of C, there exist a sequence $x_n\in C$ such that $x_n \to m$, and you have $f(x_n) = 0$, so, by continuity of $f$, you get $f(m) = f(\lim x_n) = \lim f(x_n) = 0$

Yeap, that's true. I realized that after a couple of minutes I posted the question. Sorry man. But for the second part, I'm thinking about mimic your suggestion above. I wonder what if we take the first branch of $f(x)$ only works for rational $d(x, C)$ when $d(x, C)\geq \sqrt{d(m,C)}$. The value of $f(x) = 0$ at all irrational when $d(x,C)\geq \sqrt{d(m,C)}$. I don't know how to define the other branch of $f(x)$ to make it a continuous function though:P

can you please help with the second part in case you got it?

I gave you an exemple for the second part

Many thanks for your help, Tryss! I thought about your example for part (b), but not too sure on 3 following points: (1) why X countable is relevant to the countable number of $\lambda$ outside of$ [0,1]$? (2) $\lambda$ in $f(x)$ definition should be $\lambda_{0}$? But then how come $f(x)$ is still continuous, since if I choose a point $x$ in such a way that lim $d(m,x)$ approaches $\lambda_{0} d(m,C)$ from both sides, then $f(x)$ has a jump discontinuity at that point? (3) how can the fact that C is closed be used now?

1) If X was not countable, there is no way to assume the existence of the $\lambda_0$, that is fundamental to the continuity of f. 2) yes, it's $\lambda_0$, and the trick is : there is no points in the middle : both sets are open ! 3) it's used in the same way : it garantee the fact that d(x,C) > 0

it's quite clear to me now except the parts that "both sets are open." Can you please help show the continuity part to elaborate on this point?

A simple way to show the continuity, with the definition of continuity in a topological space: $d(x) = d(x,m)$ is a continuous function. Then $d^{-1}([0,\lambda_0d(m,C)[)$ is open as the inverse image of an open set ($[0,\lambda_0d(m,C)[$ is open in $[0,+\infty[$). But $$d^{-1}(\{0\}) = \{x : d(x,m) < \lambda_0 d(m,C) \} = \{x : f(x)=0 \} = f^{-1}(\{0\})$$. Hence $f^{-1}(\{0\})$ is open. Then you show in the same way that $f^{-1}(\{1\})$ is open, and because that's the only non trivial open in $\{0,1\}$, the fonction is continuous

Another way, more direct : if $f(x) = 1$, then $d(x,m) = c > \lambda = \lambda_0d(m,C)$. Then you consider the ball $B(x,\frac{c-\lambda}{3})$, and it's easy to show that every point $y$ in this ball verify $$d(y,m) \geq d(x,m)-d(y,x) \geq c - \frac{c-\lambda}{3} > \lambda$$. Hence $f(y) = 1$. Same idea for the points where $f(y) = 0$

what does $d^{-1}([0, \alpha_{0}d(m,C)])$ mean? I think we can use the same trick you used to prove the existence of open ball such that every point $y$ in that ball gives $f(y)=1$ to obtain $f(x)=0$ for every point $x$ in the open ball $C(k, \frac{\alpha-c}{3})$? Btw, I think $\lambda_{0}$ should be in $(0,1)$ so that $f(C)=0$?

The notation is a little heavy ;) $$d^{-1}( [0,\lambda_0 d(m,C) [ ) = \left\lbrace x \in X : d(x) \in [0,\lambda_0 d(m,C) [ \right\rbrace$$ or in other words, $$d^{-1}( [0,\lambda_0 d(m,C) [ ) = \left\lbrace x \in X : 0\leq d(x,m) < \lambda_0 d(m,C) \right\rbrace$$

Did you mean there exists countable $\lambda_0$ in $[0,1]$ in the first post?

There exist an uncountable number of lambda_0 such that {x : d(x,m) = lambda_0} is empty

So, the idea is that you can cut X in two without touching an element

Take Q, you can cut it in two : {x : x²<2} and {x:x²>2}

both sets are open, and create a partition of Q

it's not possible to do this with R, because there is no "holes"