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

$A_1=\{ \text {closed unit disk in plane}\}$ $A_2=\{(1,y):y\in \mathbb{R}\}$ $A_3=\{(0,2)\}$ We need to confirm: there exist always a continuous real valued function $f$ on $\mathbb{R}^2$ such that $f(x)=a_j $ for $x\in A_j$ $j=1,2,3$

$1$. Iff atleast two of these number are equal.

$2$. all are equal.

$3$. for all values of these 3 numbers.

$4$.iff $a_1=a_2$

Is some how I need to use Urysohn's lemma here?

share|cite|improve this question
Since $A_1\cap A_2\ne\emptyset$, you need $a_1=a_2$. (Even if you don't require the function to be continuous.) – Martin Sleziak Jun 13 '12 at 11:35
which theory or result we are using here could you please tell me? – Un Chien Andalou Jun 13 '12 at 11:36
Too see this, you only need the definition of function. (I am assuming that by closed unit disc you mean $A_1=\{(x,y)\in\mathbb R^2; x^2+y^2\le 1\}$.) – Martin Sleziak Jun 13 '12 at 11:37
ya thats true.... – Un Chien Andalou Jun 13 '12 at 11:59
@Mex No, you do not need Urysohn's lemma. In each case try to construct such a function or prove that no such function can exist. – AD. Jun 13 '12 at 12:35
up vote 4 down vote accepted

As discussed in the comments you need to have $a_1=a_2$ (note that $(1,0)\in A_1\cap A_2$), so the only options are 2) and 4), where 2) is the stronger assumption. We can show however that 4) suffices.

Indeed we may apply Urysohn to the sets $A=A_1\cup A_2$ and $B=A_3$. Both sets are closed and disjoint, moreover $\mathbb R^2$ is normal. If $a_1=a_3$ choose $f\equiv a_1$. So assume $a_1\neq a_3$.

By Urysohn there exists a continuous function $f:\mathbb R^2\to [0,1]$ with $f(A)=0$ and $f(B)=1$. Postcompose this map with the canonical homeomorphism $[0,1]\to [a_1,a_3]$ if $a_1< a_3$ or the strictly decreasing homeomorphism $[0,1]\to [a_3,a_1]$ if $a_3< a_1$. We are done.

Edit: Maybe this is actually a bit of an overkill. Since your sets are given explicitely you can just define $f(x,y)=\begin{cases}a_1& \text{ if $x\leq 1$ }\\ a_3&\text{ if $x\geq 2$ }\\ a_1+(x-1)(a_3-a_1)&\text{ if $1\leq x\leq 2$ }\end{cases}$

share|cite|improve this answer
No, 2) is ok and you to apply Urysohn – AD. Jun 13 '12 at 12:38
Sorry, I don't know what you mean, can you please clarify? – Simon Markett Jun 13 '12 at 12:39
1. If $a_1=a_2=a_3$ then we may choose $f$ as a constant function. – AD. Jun 13 '12 at 12:41
I like the updated answer much more. Note also that "Urysohn for metric spaces" is much easier than the full lemma. – Dylan Moreland Jun 13 '12 at 12:46
No, $a_1$ and $a_2$ have to be equal, otherwise it doesn't work for points in the (non-trivial) intersection $A_1\cap A_2$. Therefore 1) and 3) are to weak. 4) Is true, but stronger than 2). 2) is sharp – Simon Markett Jun 13 '12 at 12:47

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.