1
$\begingroup$

Let $B := \{(x, y) \in \mathbb{R}^2 : x^2 + y^2 \le 1\}$be the closed ball in $\mathbb{R^2}$ with center at the origin. Let I denote the unit interval $[0, 1].$ Which of the following statements are true?

Which of the following statements are true?

$(a)$ There exists a continuous function $f : B \rightarrow \mathbb{R}$ which is one-one

$(b)$ There exists a continuous function $f : B \rightarrow \mathbb{R}$ which is onto.

$(c)$ There exists a continuous function $f : B \rightarrow I × I$ which is one-one.

$(d)$ There exists a continuous function $f : B \rightarrow I × I$ which is onto.

I thinks none of option will be correct

option $a)$ and option $b)$ is false Just using the logics of compactness, that is $\mathbb{R}$ is not compacts option c) and option d) is false just using the logic of connectedness that is $B-\{0\}$ is not connected but $I × I-\{0\}$ is connectedness

Is my logics is correct or not ?

Any hints/solution will be appreciated

thanks u

$\endgroup$
8
  • 1
    $\begingroup$ why does compactness of $B$ help in (a)? $f$ is not onto there. Also, $B-\{0\}$ is certainly connected. $\endgroup$
    – Randall
    Jan 11 '19 at 17:43
  • $\begingroup$ @Randall B is a circle , cut the circle it will disconnect $\endgroup$
    – jasmine
    Jan 11 '19 at 17:45
  • 2
    $\begingroup$ I have no idea what you're saying. $B$ is a solid disk. If you poke a hole in a disk it is still connected. $\endgroup$
    – Randall
    Jan 11 '19 at 17:46
  • 1
    $\begingroup$ I don't know.... $\endgroup$
    – Randall
    Jan 11 '19 at 17:52
  • 1
    $\begingroup$ @Randall compactness does help if you know dimension theory too: if $f: B \to \mathbb{R}$ were continuous and 1-1, $f[B]$ would be homeomorphic to $B$ by compactness. But $\dim f[B] \le 1$ while $\dim B=2$. Bit overkill though. $\endgroup$ Jan 11 '19 at 18:41
1
$\begingroup$

Option a) is false because we cannot even find such a map from $S^1$, the unit circle by the simplest version of Borsuk-Ulam: there are already points $x$ and $-x$ on the boundary of $B$ that have the same value.

Option b) is indeed most easily disproved by noting that $f[B]$ is compact and the reals are not.

Options c) and d) are true: $B$ is homeomorphic to $ I \times I$, as is well-known. A homeomorphism will fulfill both. Note that $B\setminus\{0\}$ is actually connected so your proposed argument doesn’t work.

$\endgroup$
1
$\begingroup$

(a) is false

As $B$ is a compact and $f$ is continuous, $f$ is an homeomorphism from the compact $B$ to $f[B]$. As $B$ is connected, $f[B]$ is also connected and is therefore an interval. However $B\setminus \{0\}$ is connected and $f[B\setminus \{0\}]$ cannot be connected. That can’t be as $f$ is an homeomorphism.

(b) is false

The image of the compact $B$ must be compact and $\mathbb R$ isn’t.

(c) and (d) are true

Consider the application that transform a ray of the unit ball into the line segment joining the origin of $B$ to the point of square aligned with the original ray.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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