Let be $f:I\to\mathbb{R}$ a continuous function, $[x,x_0]\subseteq I$ and $M:=\sup\{f(y)\mid y\in[x,x_0]\}$. Show that the supremum attains the value of $f(x_0)$ when $x\to x_0$.
My original idea was to argue with the rules of limits:
By definition of the supremum we know that for each $\epsilon>0$ there exists a $y\in [x,x_0]$ such that $f(y)<M<f(y)+\epsilon$. Then $x\to x_0$ leads to
$$\implies y\to x_0\implies\lim\limits_{y\to x_0}f(y)=f(x_0)\\\implies f(x_0)<M<f(x_0)+\epsilon \implies M=f(x_0),$$
as we can make $\epsilon>0$ arbitrarily small.
However, when I submitted this answer my tutor put a comment saying that this conclusion is not completely wrong (and maybe intuitively right) but not quite rigorous. Further he asked which specific rule would allow me to do that and finally deducted points.
Second attempt with $\epsilon$–$\delta$-criterion:
Again by definition of the supremum we know that for each $\epsilon>0$ there exists a $y\in [x,x_0]$ such that $f(y)<M<f(y)+\epsilon$. So if I now consider $x\to x_0$ then for an arbitrarily small $\epsilon>0$ there exists a $\delta>0$ such that for all $x$ with $|x-x_0|<\delta\implies |f(x)-f(x_0)|<\epsilon$. As $y\in [x,x_0]$ it follows $f(x_0)-\epsilon<f(y)<M<f(y)<f(x_0)+\epsilon$. Hence $M=f(x_0)$.
I don't understand what the problem was with my first attempt? Is my second one right?
EDIT:
Beside my both approaches I am wondering if I am able to simply argue $\lim\limits_{x\to x_0}M=\sup\{f(y)\mid y\in[x,x_0]\}=\sup\{f(y)\mid y\in[x_0,x_0]\}=f(x_0)$ because if I shrink the interval $[x,x_0]$ then $x_0$ is the only value that's left. However in this approach continuity doesn't play a role which seems strange!?
Best Answer
The problem in your first attempt is that, $y$ depends on $\varepsilon .$ Roughly speaking, you can regard this dependence as a one-to-one correspondence(N.B. this saying may not be logically correct because some $y$ may corresponds to many different $\varepsilon .$ But what I want to say is that, when you change $y$, $\varepsilon$ may be changed too.) So, when you send $y$ to $x_0$, $\varepsilon$ may change too. And it's quite hard to manage two changes simultaneously. And actually the real problem is that, you need to fix $\varepsilon$ first when $y\to x_0$, but this can not be done as what I just illustrated.
While in your second attempt, you send $\varepsilon$ to $0$, which forces $f(y)=f(x_0)$ by $$ f(x_0)-\epsilon<f(y)<M<f(y)+\varepsilon<f(x_0)+2\epsilon .$$
The reason why you can let $\varepsilon \to 0^+$ is that $x\to x_0,$ which promises that $|x-x_0|<\delta$ for any $\delta >0.$
You don't need $f(y)$ is fixed when $\varepsilon \to 0^+$.
So this is just the essential difference between your first attempt and the second one.