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About the preimage of a compact set

  1. Oct 21, 2007 #1
    If f from R to R is continuous, does it then follow that the pre-image of the closed unit interval [0,1] is compact?

    -At first I thought of a counterexample like f=sinx but it seems that its range is not R. So will the answer be yes? And how can we prove it? Will the preimage have to be bounded in this case?

  2. jcsd
  3. Oct 21, 2007 #2
    false, take f=1 for all x in R.
    then preimage of the closed unit interval is R which is not compact.
  4. Oct 21, 2007 #3

    but the range of this f is not R, is it?

  5. Oct 21, 2007 #4


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    The range of a function is not necessarily the same as its image.

    (But it's easy enough to modify yours, or SiddharthM's, suggestion so that the image of the function really is all of R)
  6. Oct 21, 2007 #5
    you mean if f is ONTO R?. how about f(x)=xsinx, the preimage of the closed unit interval here isn't bounded! So by heineborel it isn't compact.
  7. Oct 21, 2007 #6
    great! thanks...

  8. Oct 21, 2007 #7
    thank you. but do we usually take Rn as the domain and Rm as the range when we say "f is from Rn to Rm"? I am a bit confused about this...

  9. Oct 21, 2007 #8


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    f(x) = 0 is a perfectly good definition for a function from R to R.

    But the image of this function is the set {0}.
  10. Oct 21, 2007 #9
    Thanks for your patience:). But the following example tells my confusion...

    Perhaps we are familiar with a statement like this:"if f:Rn to Rm is continuous and B in Rn is bounded, then f(B) is Rm is bounded". We know that boundedness is not preserved under continuous mapping. But this statement is true in that Rn is its domain... So I am confused about the exact meaning "f is Rn to Rm".

  11. Oct 21, 2007 #10


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    There exist continuous functions that don't preserve boundedness. That doesn't imply all continuous functions don't preserve boundedness.

    In the case you cited, the theorem follows from the fact that continuous maps preserve compactness. (And the properties of Euclidean space)

    I'm not sure why you think this relates to the notion of range, though.
  12. Oct 21, 2007 #11
    when i wrote down the previous post, i was worried whether i made my question clear:)

    The statement is true because we exclude cases like: f(x)=tanx, x in (-pi/2, pi/2). We can exclude such cases as we require the domain of f(x) is Rn.

    So my question becomes: if we say "f is from Rn to Rm", do we mean "the domain of f is Rn but the range of f is a subset of Rm"?

    Thanks again:)
  13. Oct 21, 2007 #12


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    I have seen the word "range" used in two different ways:
    (1) The target of a function. (Rm, for your example)
    (2) The image of a function. (f(Rn), in your example)

    It appears to me like your issue is entirely due to mixing up the two usages of the word "range".

    (the target is also called the codomain)
  14. Oct 22, 2007 #13
    look at the definition of a ONTO or SURJECTIVE function (these two capitalized words mean the same thing). When a function's domain and range are specified it only means that a function is a rule that assigns to each value in the domain ONE value from the range. The IMAGE of the ENTIRE domain can still be a proper subset of the range. A ONTO function is one where the image of the domain is the ENTIRE range.
  15. Oct 22, 2007 #14
    i almost get it:) Thanks.
  16. Oct 23, 2007 #15


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    you are stumbling around over the trivial part of the topic, i.e. the domain and range.

    the interesting part is: when is the preimage of a compact set compact? continuos functions with this property are called proper and are very important.
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