Which of the following statements are true or false? Give proofs or counterexamples to justify your answers.I said that this statement was false and used the counterexample $\displaystyle g(x)=x$ and $\displaystyle f(x)= \begin{cases} x \ \ \ \ \ x \in \mathbb{Q}\\ 0 \ \ \ \ \ x \in \mathbb{R}- \mathbb{Q} \end{cases}$i). If $\displaystyle g: \mathbb{R} \rightarrow \mathbb{R}$ is continuous and $\displaystyle f: \mathbb{R} \rightarrow \mathbb{R}$ is such that $\displaystyle f(x)=g(x) \ \forall \ x \in \mathbb{Q}$, then f is continuous.

I put true.ii). If $\displaystyle f: \mathbb{R} \rightarrow \mathbb{R}$ and $\displaystyle g: \mathbb{R} \rightarrow \mathbb{R}$ are both continuous and $\displaystyle f(x)=g(x) \ \forall \ x \in \mathbb{Q}$, then f=g.

Since g is continuous I know that $\displaystyle \forall \ \epsilon_1>0 \exists \delta_1>0 \ s.t \ |x-c|< \delta \Rightarrow |f(x)-f(c)|< \epsilon_1$.

f is also continuous so $\displaystyle \forall \ \epsilon_2>0 \exists \delta_2>0 \ s.t \ |x-c|< \delta \Rightarrow |g(x)-g(c)|< \epsilon_2$.

$\displaystyle f(c)-\epsilon_1<f(x)<f(c)+\epsilon_1$

$\displaystyle g(c)-\epsilon_2<g(x)<g(c)+\epsilon_2$

$\displaystyle -|f(c)-g(c)|-\epsilon_1+\epsilon_2 \leq f(c)-\epsilon_1-g(c)+\epsilon_2<f(x)-g(x)<f(c)-g(c)+\epsilon_1-\epsilon_2 $$\displaystyle \leq |f(c)-g(c)|+\epsilon_1-\epsilon_2$

Let $\displaystyle \epsilon_3=|f(c)-g(c)|$

$\displaystyle -\epsilon_3-\epsilon_1+\epsilon_2<f(x)-g(x)<\epsilon_3+\epsilon_1-\epsilon_2$

Define $\displaystyle \epsilon:= \epsilon_1-\epsilon_2+\epsilon_3$. WLOG, let $\displaystyle \epsilon_1+\epsilon_3> \epsilon_2$.is this step valid?

This gives $\displaystyle |f(x)-g(x)|< \epsilon$ as required.

Since my initial definitions only apply to a rational c, would I need to prove they are equal when c is not rational?

I opted for true.iii). If $\displaystyle f: \mathbb{R} \rightarrow \mathbb{R}$ is continuous at 0 with $\displaystyle f(0)=0$ and $\displaystyle g: \mathbb{R} \rightarrow \mathbb{R}$ is bounded then the product fg is continuous at 0.

f is continuous:

$\displaystyle \forall \ \epsilon>0 \ \exists \ \delta>0 \ s.t \ |x|< \delta \Rightarrow |f(x)-f(0)|< \epsilon$.

g bounded:

$\displaystyle \exists \ C \in \mathbb{R} \ s.t \ -C<g(x)<C$ where $\displaystyle C= max \{|lower \ bound \ of \ g(x)|, \ |upper \ bound \ of \ g(x)| \}$.

$\displaystyle f(0)-\epsilon<f(x)<f(0)+\epsilon$

$\displaystyle -C<g(x)<C$

$\displaystyle \Rightarrow \ -C(f(0)-\epsilon)<f(x)g(x)< C(f(0)+ \epsilon)$

Let $\displaystyle \epsilon_1=max \{ |C(f(0)- \epsilon)|, \ |C(f(0)+ \epsilon)| \}$

$\displaystyle \Rightarrow - \epsilon_1<f(x)g(x)< \epsilon_1$

Since I am only interested in the behaviour of the multiple functions at 0, pick $\displaystyle \delta_1:=\epsilon_1$.can you do this? I need to restrict delta somehow so large values of delta can also be used.

This gives that:

$\displaystyle \forall \ \epsilon_1>0 \ \exists \ \delta_1>0 \ s.t \ |x|< \delta_! \Rightarrow |f(x)g(x)-f(0)g(0)|< \epsilon_1$ as required.

I said this is true since the composition of two continuous functions is itself continuous. However, I cannot formulate a proof!If $\displaystyle f:[-1,1] \rightarrow \mathbb{R}$ is such that the composition $\displaystyle x \rightarrow \sin f(x)$ (this is a mapping, not a limit) is continuous on $\displaystyle \mathbb{R}$ then f is continuous.

I have a feeling this is false. Unfortunately I am unable to think of a counterexample.If $\displaystyle f:[-1,1] \rightarrow \mathbb{R}$ is such that the composition $\displaystyle x \rightarrow f(\sin x)$ (mapping once again) is continuous on $\displaystyle \mathbb{R}$ then f is continuous.

Any help with any of these questions would be appreciated. They get so much harder towards the end!