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Math Help - Continuous functions: Images and Pre-images of closed and open sets

  1. #1
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    Continuous functions: Images and Pre-images of closed and open sets

    I'm having a bit of difficulty seeing whether a function or its inverse preserves whether a set's image/pre-image remains open of closed. There are 5 parts for which I must either prove or give a counter-example. I cannot find an example for 3. and can't think of how to prove 4. I put all of them down though - just in case I messed anything up.

    Setup:
    Let  f:\mathbb{R} \to \mathbb{R} be continuous. Also, let S \subseteq \mathbb{R} and T \subseteq \mathbb{R}.

    1. If S is open, then f(S) is open.
    False: If f(x) = c for some constant c, then its range is closed no matter the pre-image.

    2. If S is closed, then f(S) is closed.
    False: Using f(x)=e^x and S = ( - {\infty} , 0 ), then f(S) = (0,1).

    3. If S is bounded, then f(S) is bounded.
    False: But I'm having difficulty thinking of an example.

    4. If T is open, then f^-1 (T) is open.
    True: But again, I'm not sure where to turn for the proof.

    5. If T is closed, then f^-1 (T) is closed.
    True: If I had the above, I'd use its complement to show this one holds, but since I don't... Let x \in f^-1 (T) and let x_n be a sequence of elements from x \in f^-1 (T) that converges to x. Since f is continuous, f(x_n) converges to f(x). Since C is closed, f(x) is in C. So, x is in f^-1 (C) .
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  2. #2
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    Re: Continuous functions: Images and Pre-images of closed and open sets

    I realize I can use the complement of 5. to get 4. - I would, however, still appreciate a direct method if anyone knows.

    As for 3., I might be going crazy .. intuitively, I think the statement is true. After all, since f is continuous on \mathbb{R}, it is continuous on S. If it were an interval l could look at [a,b], where a=inf(S) and b=sup(S), I could use continuity to show that the function is bounded on the interval. But S is a set - and I'm not sure how that changes things.
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  3. #3
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    Re: Continuous functions: Images and Pre-images of closed and open sets

    Quote Originally Posted by jsndacruz View Post
    Setup:
    Let  f:\mathbb{R} \to \mathbb{R} be continuous. Also, let S \subseteq \mathbb{R} and T \subseteq \mathbb{R}.

    1. If S is open, then f(S) is open.
    False: If f(x) = c for some constant c, then its range is closed no matter the pre-image.

    2. If S is closed, then f(S) is closed.
    False: Using f(x)=e^x and S = ( - {\infty} , 0 ), then f(S) = (0,1).

    3. If S is bounded, then f(S) is bounded.
    False: But I'm having difficulty thinking of an example.

    4. If T is open, then f^-1 (T) is open.
    True: But again, I'm not sure where to turn for the proof.

    5. If T is closed, then f^-1 (T) is closed.
    True: If I had the above, I'd use its complement to show this one holds, but since I don't... Let x \in f^-1 (T) and let x_n be a sequence of elements from x \in f^-1 (T) that converges to x. Since f is continuous, f(x_==nn) converges to f(x). Since C is closed, f(x) is in C. So, x is in f^-1 (C) .
    Be careful in #2. (-\infty,0) is not close. But (-\infty,0] is closed, while (0,1] is not.


    Quote Originally Posted by jsndacruz View Post
    I realize I can use the complement of 5. to get 4. - I would, however, still appreciate a direct method if anyone knows.
    As for 3., I might be going crazy .. intuitively, I think the statement is true. After all, since f is continuous on \mathbb{R}, it is continuous on S. If it were an interval l could look at [a,b], where a=inf(S) and b=sup(S), I could use continuity to show that the function is bounded on the interval. But S is a set - and I'm not sure how that changes things.
    Every bounded set of real numbers is the subset of a closed finite interval.
    So it is the subset of a compact set. The function is continuous. Thus?
    Thanks from jsndacruz
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  4. #4
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    Re: Continuous functions: Images and Pre-images of closed and open sets

    Ah - for #2 I meant to use square brackets. Thanks for paying attention to me!

    My professor never taught us about 'compact' sets - my grasp on topology is really poor. But I think I know what you're getting at: since S is a subset of a closed interval, it is sufficient to show that the function is bounded on that interval, since the subset of a bounded set must be bounded.
    Last edited by jsndacruz; April 16th 2012 at 09:18 AM. Reason: tiny little grammar things
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