Decide if the following claims are true or false, providing either a short proof or counterexample to justify each conclusion. Assume throughout that g is defined and continuous on all of R.

If \(g(r) = 0\) \(\forall r \epsilon\)Q, then \(g(x) = 0\) \(\forall x \epsilon\)R.

If \(g(r) = 0\) \(\forall r \epsilon\)Q, then \(g(x) = 0\) \(\forall x \epsilon\)R.

Recall:

Theorem 3.2.10 (Density of Q in R). For every y \(\epsilon\) R, there exist a sequence of rational numbers that converges to y.

Thus we can construct a sequence \((r_n)\) such that \((r_n) \rightarrow x\) \(\forall x \epsilon\)R by the Density of Q in R.

If \(g(r) = 0\) \(\forall r \epsilon\)Q, then \(g(x) = 0\) \(\forall x \epsilon\)R.

Thus we can construct a sequence \((r_n)\) such that \((r_n) \rightarrow x\) by the Density of Q in R.

Recall:Theorem 4.3.2 (Characterizations of Continuity). Let f : A \(\rightarrow\) R, and let \(c \epsilon A\).

1. For all \((x_n) \rightarrow c\) (with \(x_n \epsilon A\)), it follows that \(f(x_n) \rightarrow f(c)\).

Since our g is continuous we know 1 is a characteristic of g.

Thus from Theorem 4.3.2 we know that \((\forall (r_n) \rightarrow x (r_n \epsilon Q \subset R))\rightarrow (g(r_n) \rightarrow g(x))\) is true.

If \(g(r) = 0\) \(\forall r \epsilon\)Q, then \(g(x) = 0\) \(\forall x \epsilon\)R.

Thus we can construct a sequence \((r_n)\) such that \((r_n) \rightarrow x\) by the Density of Q in R.

Thus from Theorem 4.3.2 we know that \((\forall (r_n) \rightarrow x (r_n \epsilon Q \subset R))\rightarrow (g(r_n) \rightarrow g(x))\) is true.

\(\because\) all \(r_n\) are rational numbers we know that \(g(r_n)=0\).

If \(g(r) = 0\) \(\forall r \epsilon\)Q, then \(g(x) = 0\) \(\forall x \epsilon\)R.

Thus we can construct a sequence \((r_n)\) such that \((r_n) \rightarrow x\) by the Density of Q in R.

Thus from Theorem 4.3.2 we know that \((\forall (r_n) \rightarrow x (r_n \epsilon Q \subset R))\rightarrow (g(r_n) \rightarrow g(x))\) is true.

\(\because\) all \(r_n\) are rational numbers we know that \(g(r_n)=0\).

\(\therefore g(x)=0\) \(\forall x \epsilon\)R.

Recall:

Decide if the following claims are true or false, providing either a short proof or counterexample to justify each conclusion. Assume throughout that g is defined and continuous on all of R.

If \(g(x_0)>0\) for a single point \(x_0 \epsilon R\), then \(g(x)\) is in fact strictly positive for uncountably many points.

If \(g(x_0)>0\) for a single point \(x_0 \epsilon R\), then \(g(x)\) is in fact strictly positive for uncountably many points.

let \(\varepsilon = \frac{1}{2} g(x_0)\). Note that \(V_\varepsilon(g(x_0)) > 0\) \(\forall y \epsilon V_\varepsilon(g(x_0))\).

If \(g(x_0)>0\) for a single point \(x_0 \epsilon R\), then \(g(x)\) is in fact strictly positive for uncountably many points.

let \(\varepsilon = \frac{1}{2} g(x_0) >0\).

Note that \(V_\varepsilon(g(x_0)) > 0\) \(\forall y \epsilon V_\varepsilon(g(x_0))\).

Because g is continuous we know that for this \(\varepsilon\) it is true that there exist a \(\delta\) such that whenever \(|x-x_0|<\delta\) it follows that \(|g(x)-g(x_0)|<\varepsilon\). That is, there exist a \(V_\delta(x_0)\) such that \(V_\varepsilon(g(x_0))>0\).

If \(g(x_0)>0\) for a single point \(x_0 \epsilon R\), then \(g(x)\) is in fact strictly positive for uncountably many points.

let \(\varepsilon = \frac{1}{2} g(x_0) >0\).

Note that \(V_\varepsilon(g(x_0)) > 0\) \(\forall y \epsilon V_\varepsilon(g(x_0))\).

Because g is continuous we know that for this \(\varepsilon\) it is true that there exist a \(\delta\) such that whenever \(|x-x_0|<\delta\) it follows that \(|g(x)-g(x_0)|<\varepsilon\). That is, there exist a \(V_\delta(x_0)\) such that \(V_\varepsilon(g(x_0))>0\).

\(\because V_\varepsilon(g(x_0))\) is an open interval on R, It has the same cardinality as R. Thus, If \(g(x_0)>0\) for a single point, then \(g(x)\) is in fact strictly positive for uncountably many points.