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I tried to use the Cantor-Bernstein Theorem. First, we have $S\subset S\cup T$, so that $\left | S \right |\leqslant \left | S\cup T\right | $. This implies $\left | S\cup T \right |\geqslant c$. But I don't know how to prove the opposite direction. Could someone help me with it? Thank you very much!

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Hint: $S=_c[0,1], T=_c[2,3]{}$.

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  • $\begingroup$ Thanks for answering. Do you mean that S∪T ⊂ R, so that ∣S∪T∣⩽ ∣R∣=c. I don't understand that since c is a finite number, how could S, T has infinite elements? $\endgroup$ – Catiger3331 Mar 19 '14 at 1:11
  • $\begingroup$ @LiangkaiHu I do mean that $|S\cup T|\leq |\mathbb R|=\mathfrak c$.But I don't get what you mean with $\mathfrak c$ being a finite number. $\endgroup$ – Git Gud Mar 19 '14 at 12:12

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