I am solving previous years problems in analysis/ topology of my institute and I am doubtful about this answer of this question. Question is - Can there exists a continuous function from [-1,1] to R ( Set of Real numbers) which is onto. I think there can exists as any interval is equivalent ( cardinality wise) to set of real numbers. But the answer is no. Please help.
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2Note: equicardinality does not require a continuous function – J. W. Tanner Aug 05 '19 at 17:31
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One can show that the continuous image of a compact set is again compact, see for example https://math.stackexchange.com/a/226328/450140. As $\mathbb{R}$ is not compact with the standard topology this proves that such map can not exist.
Jonas Lenz
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A theorem in analysis states that a continuous function on a compact set (e.g. closed interval) is bounded and attains its minimum and maximum. Because $\mathbb{R}$ is unbounded, such a function cannot map to all of $\mathbb{R}$.
D.B.
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