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f(x) = x^{2} is a continuous function on the set R of real numbers, and (-1, 1) is an open set in R, but f(-1, 1) = [0, 1), and [0, 1) is not an open set in R. So, f is not an open function on R.

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Q: Is the image of an open set under a continuous mapping need not be open?
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