4.4: 8, 11, 12Proof. It is considered impossible. Becausefis continuous and [0,1] is compact, byTheorem 4.4.1, the range is compact as well. However, (0,1) is not considered compact.Better said, there cannot exist a continuous function defined on [0,1] that possesses somerange (0,1). Since [0,1] is a compact set and continuous image of the compact srt [0,1]will be compact while the range set (0,1) is not compact. QED.Proof. Letf: (0,1), which would be given byf(x) =+sinwith→ [0, 1]1212(2π?). So, the functionfis continuous and onto. Therefore, the set (0,1) has some? ∈ (0, 1)range [0,1] under such a map as mentioned above. QED.Proof. There cannot be some continuous function from the set (0,1] with range set (0,1).Now, if it is possible, we let, which implies that?((0, 1] )= (0, 1)?−1(0, 1)=(0, 1] .This can be considered a contradiction, sinceis open asfa continuous map, but?−1(0, 1)(0,1] is not open. QED.Proof. () Let us make the assumption thatis indeed continuous onRand letRbe⇒?𝑂 ⊆open. We yearn to show thatis open. Then, we fixand demonstrate?