Incomplete question:
Let us take an example related to given statement:
Suppose A is bounded and B={x2 | x is a member of A}. Show if sup(A)=α \alpha α then sup(B) = α \alpha α
Solution:
Given that A is bounded sub set of R \mathbb{R} R and sup A = α \sup A=\alpha sup A = α and B = { x 2 / x ∈ A } B=\left\{x^{2} / x \in A\right\} B = { x 2 / x ∈ A }
Since x ≤ α ∀ x ∈ A x \leq \alpha \forall x \in A x ≤ α ∀ x ∈ A
x 2 ≤ α 2 x 2 ≤ α 2 ∀ x 2 ∈ B \begin{aligned}
&x^{2} \leq \alpha^{2} \\
&x^{2} \leq \alpha^{2} \forall x^{2} \in B
\end{aligned} x 2 ≤ α 2 x 2 ≤ α 2 ∀ x 2 ∈ B
Therefore α 2 \alpha^{2} α 2 is upper bound for B
Let β \beta β be an other upper bound for B
so that x 2 ≤ β ∀ x 2 ∈ B x^{2} \leq \beta \forall x^{2} \in B x 2 ≤ β ∀ x 2 ∈ B
⇒ x ≤ β ∀ x ∈ A \Rightarrow x \leq \sqrt{\beta} \forall x \in A ⇒ x ≤ β ∀ x ∈ A
Therefore β \sqrt{\beta} β is upper bound for A.
Since α \alpha α is suprimum of A, we have α ≤ β ⇒ α 2 ≤ β \alpha \leq \sqrt{\beta} \Rightarrow \alpha^{2} \leq \beta α ≤ β ⇒ α 2 ≤ β .
Therefore α 2 \alpha^{2} α 2 is least upper bound for B
Hence sup B = α 2 =\alpha^{2} = α 2
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