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Suppose $f : mathbbR^n rightarrow mathbbR$ is continuous and
$f(u) > 0$ if the point $u in mathbbR^n$ has at least one
rational component. Prove that $f(u) geq 0 $ for all $u in
mathbbR^n$.

Partial Solution:

What remains to be proven is that $f(u) geq 0$ for every $u in mathbbR^n$ with at all irrational components. Let $v$ be such a vector. Since every real number is the limit point of a rational sequence, we can define sequences $r_1, r_2, ldots, r_n,$ where $r_k$ converges to the $k^textth$ component of $v$.

I don't really know how to finish from here. Can someone please help me?

For fixed $v in mathbbR^n$ write $v=(v^(1),ldots,v^(n))$, i.e. $v^(k)$ is the $k$-th component of $v$ for $k in 1,ldots,n$.

As you suggest, we choose for some some $k in 1,ldots,n$ a rational sequence $(r_i)_i in mathbbN$ such that $r_i$ converges to $v^(k)$ as $i to infty$. Now define $$u_i := (v^(1),ldots,v^(k-1),r_i, v^(k+1),ldots,v^(n))$$ for $i geq 1$. By construction, we have $f(u_i) geq 0$ for all $i$ and $u_i to v$ as $i to infty$.

Hint: Use the continuity of $f$ to conclude that $f(v) geq 0$. (How can you express $f(v)$ in terms of $f(u_i)$?)

Let's try to prove a stronger result:

Theorem. Suppose $f : mathbbR^n rightarrow mathbbR$ is continuous and
$f(u) > 0$ if the point $u in mathbbR^n$ has all its components rational. Then $f(u) geq 0 $ for all $u in
mathbbR^n$.

Proof. Let $v=(v_1,v_2,...,v_n)inmathbbR^n$. Choose rational sequences $x_mk$ such that $lim_kto inftyx_mk=v_m$, for all $m=1,2,...,n$. The sequence of points $x_k=(x_1k,...,x_nk)$. Please tell me if you want to know in details why such choice is possible.

Then $x_k to v$. Since $f$ is continuous, $f(x_k) to f(v)$. We know that $f(x_k)>0$ for all $k$. To show that $f(v)geq0$, suppose that $f(v)<0$. Then for all $k$, $|f(x_k)-f(v)|geq|f(v)|$. This is a contradiction, because for $f(x_k) to f(v)$, there should exist $k_0$ such that for any $kgeq k_0$, $|f(x_k)-f(v)|leq|f(v)|$. Here, we take $|f(v)|>0$ as the arbitrary number $epsilon>0$.

Note. The strict inequality becomes NOT strict after taking the limit. We can actually find examples to show that equality can hold in $f(u) geq 0 $.

Hint: Every $uinBbbR^n$ is the limit of a sequence $(u_k)_k$ where each of the $u_k$ has at least one rational component, and hence $f(u)=lim_ktoinftyf(u_k)geq0$.