Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top


  1. There is a continuous onto function from the unit sphere in $\mathbb{R}^3$ to the complex plane $\mathbb{C}$.

  2. There is a non-constant continuous function from the open unit disc $D = \{z ∈ \mathbb{C} \mid |z| < 1 \}$ to $\mathbb{R}$ which takes only irrational values .

  3. $f \colon \mathbb{C} \to \mathbb{C}$ is an entire function such that the function $g(z)$ given by $g(z) = f( 1/z)$ has a pole at 0. Then f is a surjective map.

please help anyone.

share|cite|improve this question
That's three questions … Some hints: 1. compactness, 2. connectedness, 3. polynomial. – Harald Hanche-Olsen Sep 19 '12 at 16:52
i did not understand the hint of the third problem. please explain briefly.thank you – mintu Sep 19 '12 at 17:17
@mintu: Look at something like $f(z)=e^z$, so that $g$ has an essential singularity at $z=0$. Recall that a function has an essential singularity iff it has infinitely many $z^{-1}$ power terms. In other words, for which $f$ can $g$ really have a pole at 0? – Alex R. Sep 19 '12 at 18:10

(1) False. The continuous image of a compact space is compact, and a compact subset of the plane is bounded. (2) False. The continuous image of a connected set is connected, and the connected subsets of $\mathbb{R}$ are intervals. (3) True. An entire function with a pole and infinite is a polynomial. Now apply the fundamental theorem of algebra.

share|cite|improve this answer

The image of a compact space under a continuous function is compact. That answers your first question if you know the plane is not compact and the sphere is. The sphere is a closed and bounded subset of a Euclidean space, so it's compact. It's closed because it's the inverse image of the closed set $\{1\}$ under the continuous function $(x,y,z)\mapsto x^2+y^2+z^2$.

For your second question, a restriction of your function on the disk to some line segment within the disk is also continuous, and you can apply the intermediate value theorem and the fact that between every two real numbers there is a rational number. Make the ends of the line segment two points that don't have the same image under the function. You know those exist since it's non-constant.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.