$\inf_{x\in[a,b]}f(x)=\inf_{x\in[a,b]\cap\mathbb{Q}}f(x)$ for a continuous function $f:[a,b]\to\mathbb{R}$ Let $f:[a,b]\to\mathbb{R}$ be continuous. I'm sure it's not hard, but I'm unsure what exactly we need to do to prove $$\inf_{x\in[a,b]}f(x)=\inf_{x\in[a,b]\cap\mathbb{Q}}f(x)$$
 A: I will give a complete proof. First, since $$[a,b]\cap 
\mathbb{Q}
\subset \lbrack a,b]$$ then $$\inf_{x\in \lbrack a,b]}f(x)\leq \inf_{x\in
\lbrack a,b]\cap 
\mathbb{Q}
%EndExpansion
}f(x).$$ Let us prove the other inequality. Because of continuity of $f$ on $%
[a,b]$ then there exists some $x_{1}\in \lbrack a,b]$ such that $$%
f(x_{1})=\inf_{x\in \lbrack a,b]}f(x).$$ Density of $[a,b]\cap 
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
$ into $[a,b]$ implies that there exists a sequence $(x_{n})_{n\geq 2}$ such
that $$\lim_{n\rightarrow \infty }x_{n}=x_{1}$$ and $$x_{n}\in 
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
$$ for any $n\geq 2.$ Therefore, $$\inf_{x\in \lbrack a,b]\cap 
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
}f(x)\leq f(x_{n})$$ for any $n\geq 2.$ Passing to the limit one gets $$%
\inf_{x\in \lbrack a,b]\cap 
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
}f(x)\leq \lim_{n\rightarrow \infty }f(x_{n})$$ and by continuity of $f$ at
the point $x_{1}$ it follows that $$\lim_{n\rightarrow \infty
}f(x_{n})=f(\lim_{n\rightarrow \infty }x_{n})=f(x_{1}).$$ That is: $$%
\inf_{x\in \lbrack a,b]\cap 
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
}f(x)\leq \inf_{x\in \lbrack a,b]}f(x).$$ Which proves the required equality
in the title.
A: Suppose that $m_r=\inf_{x\in[a,b]}f(x)$ and $m_q=\inf_{x\in[a,b]\cap\mathbb{Q}}f(x)$ are different; necessarily $m_r<m_q$.
Necessarily by continuity there exists $x_0\in[a,b]$ such that $m_q>f(x_0)>m_r$.
Let $\epsilon>0$ such that $\epsilon<m_q-f(x_0)$, then by density of $\mathbb{Q}$ in $\mathbb{R}$ and by continuity of $f$ there exists $q_0\in\mathbb{Q}$ such that $|f(q_0)-f(x_0)|<\epsilon$. That leads us to conclude that $f(q_0)<m_q$ which is absurd.
Hence $m_r=m_q$.
