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    Open set whose preimage under the given function is not open

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    Let g: R -> R by:

    g(x) = { x^2 + 2 ... if x <= 1, 5 - x ... if x > 1 }

    Find an open subset of R (w/ respect to the usual topology) whose preimage under g is not an open subset of R.

    © BrainMass Inc. brainmass.com October 10, 2019, 2:29 am ad1c9bdddf
    https://brainmass.com/math/geometry-and-topology/finding-open-subset-379964

    Solution Preview

    Consider the set S = {x: 2 < x < 3.6}. Note that S is an open subset of R with respect to the usual topology. We will prove that the preimage of S under the given function g is not an open subset of R.

    First, note the following:

    g(0) = 0^2 + 2 = 2, so 0 is not in the preimage of S.

    g(1) = 1^2 + 2 = 3, so 1 is in the preimage of S.

    2 < g(x) < 3 for every x such that 0 < x < 1

    Thus at this point we know that the preimage of S contains at least the half-open interval ...

    Solution Summary

    An example of an open subset of R whose preimage under the given function g is not open is provided, along with a complete, detailed justification that the preimage of that open set is not open.

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