Continuity and density with functions coinciding

Question

Let $(X,d)$ and $(Y,d')$ be metric spaces.

Let $f$ and $g$ be two functions with the following mapping: $X\rightarrow Y$

Let $f$, $g$ be continuous on $X$. Let $A\subseteq X$ with $A$ dense in $X$ such that $f(a)=g(a)\space\forall\space a\in A$

I want to show $f(x)=g(x)\space\forall\space x\in X$

Let $y\in X$

Since $f$ and $g$ are continuous on $X$, then $f-g$ is continuous on $X$

Suppose $f(y)>g(y)$ which is equivalent to saying $f(y)-g(y)>0$

By the definition of continuity, $\exists\space\delta_{0}>0$ such that:

If $|x-y|<\delta_{0}$, then $|(f(x)-g(x))-(f(y)-g(y))|<f(y)-g(y)$

This implies:

If $y-\delta_{0}<x<y+\delta_{0}$, then $f(x)>g(x)$

But $A$ is dense in $X$

So $\exists\space a\in (y-\delta_{0},y+\delta_{0})\cap A$

So $\exists x\in (y-\delta_{0},y+\delta_{0})$ such that $f(x)=g(x)$

We have contradiction. I can then go on to find a similar contradiction for the case of $f(y)<g(y)$ to conclude that $f(y)$ must equal $g(y)$.

Is my approach correct?

Answer 1

Your idea is not bad , if your functions where $f,g : (\mathbb{R},|.|) \to (\mathbb{R}, |.|)$ it would be correct ! But now since the space $(Y, d')$ is arbitrary you cant be sure whether you have algebraic operations in $Y$ to add and substract elemements inside $Y$.

But still you can modify your argument to prove the argument .

Suppose that there is a $y \in X$ such that $f(y) \neq g(y)$ (now you cant say $f(y) > g(y)$ or $ f(y) < g(y)$ since you dont know if there is any notion of comparison of two elements in $Y$ like we have in the well ordered set $\mathbb{R}$.)

But from $f(y) \neq g(y)$ you know that $\epsilon = d'(f(y),g(y)) >0$. (This is the same like you said $f(y)-g(y) >0$ but we cant substract $g(y) $ from $f(y)$ in the random space $Y$ ).

Now you can procceed in the same way and say :

Since $f$ is continuous at $y$ there is an $r_1>0$ such that for every $x \in X$ with $d(x,y)< r_1$ (thats the same with $|x-y| < r_1$) you have $d'(f(y),f(x)) < \frac{\epsilon}{2}.$

Again by the continuity of $g$ there is an $r_2 >0$ such that for every $x \in X$ with $d(x,y) < r_2$ you have $d'(g(x),g(y)) < \frac{\epsilon}{2}.$

Pick $r = min \{r_1,r_2\} >0$

Now since $A$ is dense there exist $\alpha \in A$ such that $ \alpha \in B(y,r)$ where $B(y,r) = \{ x \in X : d(x,y) <r\}$ , which means that $d(\alpha, y ) < r.$

Now since $d \in A$ by hypothesis $f(\alpha) = g(\alpha)$

So $\epsilon = d'(f(y),g(y)) \leq d'(f(y),f(\alpha)) + d'(g(y),f(\alpha)) =d'(f(y),f(\alpha)) + d'(g(y),g(\alpha)).$ (*)

Now remember that $d(y,\alpha) < r = min\{r_1,r_2\}$ so $d'(f(y),f(\alpha)) < \dfrac{\epsilon}{2}$ and $d'(g(y),g(\alpha))< \dfrac{\epsilon}{2}.$

So if you plug this in (*) you have $\epsilon < \epsilon$ which is a contradiction!

Hence $f(y)=g(y).$

Answer 2

No, it is not correct. For instance, what does $f(y)-g(y)$ mean?

Take $x\in X$. Then, since $A$ is dense, $x=\lim_{n\to\infty}a_n$, for some sequence of elements of $A$. But then\begin{align}f(x)&=f\left(\lim_{n\to\infty}a_n\right)\\&=\lim_{n\to\infty}f(a_n)\\&=\lim_{n\to\infty}g(a_n)\\&=g\left(\lim_{n\to\infty}a_n\right)\\&=g(x),\end{align}

Answer 3

You assume $Y$ is some set with a linear order, or a difference operation. But $Y$ is only given to be metric, and no more.

One way to go about it, is to use sequences. Suppose $x \in X$. Then as $A$ is dense, so $\overline{A}=X$, we know that there is a sequence $(a_n)$ from $A$ such that $a_n \to x$ as $n \to \infty$.

By (sequential) continuity of $f$ and $g$ and the fact that $f(a_n) = g(a_n)$ for all $n$ : $$f(x) = f(\lim_n a_n) = \lim_n f(a_n) = \lim_n g(a_n) = g(\lim_n a_n) = g(x)$$ and we use the fact that sequential limits are unique to conclude that $f(x) = g(x)$. As $x$ was arbitrary we are done.

A proof from first principles/definitions is also possible:

Suppose $f(x) \neq g(x)$ for some $x \in X$. Define $r= d'(f(x), g(x)) > 0$.

Applying the definitions of continuity at $x$ for $\epsilon = \frac{r}{2}$ we find $\delta_1 > 0$ such that

$$\forall p \in X: d(x,p) < \delta_1 \to d'(f(p),f(x)) < \frac{r}{2}$$

and $\delta_2 >0$ such that

$$\forall p \in X: d(x,p) < \delta_2 \to d'(g(p),g(x)) < \frac{r}{2}$$

Now we note that by $A$ being dense, we can find $a \in A$ such that $d(x,a) < \min(\delta_1,\delta_2)$. For this $a$ we can use both continuity conditions and so

$$d'(f(a),f(x)) < \frac{r}{2} \text{ and } d'(g(a), g(x)) < \frac{r}{2}$$

But, we also know that $f(a)=g(a)$ and so:

$$r=d(f(x), g(x)) \le d'(f(x), f(a)) + d'(f(a), g(x))= d'(f(x),f(a)) + d'(g(a),g(x)) < \frac{r}{2} + \frac{r}{2} = r$$

which is a contradiction. So $f(x) \neq g(x)$ was a false assumption, and so $f(x)= g(x)$ everywhere.