HW10 MAT127A.pdf - 4.4 8 11 12 Proof It is considered impossible Because f is continuous and[0,1 is compact by Theorem 4.4.1 the range is compact as

# HW10 MAT127A.pdf - 4.4 8 11 12 Proof It is considered...

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4.4: 8, 11, 12 Proof . It is considered impossible. Because f is continuous and [0,1] is compact, by Theorem 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 some range (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 . Let f : (0,1) , which would be given by f ( x ) = + sin with → [0, 1] 1 2 1 2 (2π?) . So, the function f is 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, since is open as f a continuous map, but ? −1 (0, 1) (0,1] is not open. QED. Proof . ( ) Let us make the assumption that is indeed continuous on R and let R be ? 𝑂 ⊆ open. We yearn to show that is open. Then, we fix and demonstrate ?