Suppose $C\subset \mathbb{R^n}$, and $f:\mathbb{R^n}\to \mathbb{R}$ is continuous on cl$C.$
Prove that $$\inf\{ f(x)|x\in C\}=\inf\{ f(x)|x\in\text{cl}C \}.$$
I tried to use whenever a sequence, $x_n \in C$ converges to $x^* \in\text{cl}C$, we have $f(x_n)\to f(x^*)$.
But I have no idea after this... I would really appreciate your help.
$\text{cl}C$ means the closure of $C$.
On
(i).If $D\subset E\subset \Bbb R$ then $\inf D\ge \inf E.$
(ii). If $D\subset \Bbb R$ then $\inf D=\inf \overline D.$
(iii). Since $f$ is continuous we have $ f(\overline C)\subset \overline {f(C)}. $
$$\text {Therefore }\quad (*)\quad\inf f(C)\ge \inf f(\overline C)\ge \inf \overline {f(C)}=\inf f(C). $$ The 1st inequality in $(*)$ is by (i); the 2nd is by (i) & (iii); the "$=$" is by (ii).
On
This could be proven by contradiction. Suppose $$\inf\{ f(x)|x\in C\}>\inf\{ f(x)|x\in\text{cl}C \}.\tag1$$ Then there is a $x^*\in\text{cl}C$ such that $$f(x^*)<\inf\{ f(x)|x\in C\}$$ Since $x^*$ is in the closure of $C$, there is a sequence $\{x_n\}\subset C$ that converge to $x^*$.
By continuity of $f$, $f(x_n)\to f(x^*)$ as $n\to\infty$. It means for every $\epsilon>0$, there exist a $N_\epsilon$ such that for every $n>N_\epsilon$ $$\lvert f(x_n)-f(x^*)\rvert<\epsilon$$ Let's take $$\epsilon=\frac{\inf\{ f(x)|x\in C\}-f(x^*)}2>0$$ For this $\epsilon$, there is a $N_\epsilon$ such that $$n>N_\epsilon\implies \lvert f(x_n)-f(x^*)\rvert<\epsilon$$ This last equation is equivalent to $$f(x^*)-\epsilon<f(x_n)<f(x^*)+\epsilon$$ By the definition of $\epsilon$ and the right inequality, we have $$f(x_n)<f(x^*)+\frac{\inf\{ f(x)|x\in C\}-f(x^*)}2=\frac{\inf\{ f(x)|x\in C\}+f(x^*)}2<\inf\{ f(x)|x\in C\}$$ Which is a contradiction since $x_n\in C$. The assumption $(1)$ is false and $$\inf\{ f(x)|x\in C\}=\inf\{ f(x)|x\in\text{cl}C \}.$$
Hint: $\inf f(\overline{C})=\inf \overline{f(\overline{C})}=\inf \overline{f(C)}=\inf f(C)$.