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Let
$B$ = {$(x, y) ∈ \mathbb{R}^2 : x^2 + y^2 ≤1$} and $D$ = {$(x, y) ∈ \mathbb{R}^2 : x^2 + y^2 < 1$}.
Pick out the true statements.
(a) Given a continuous function $g : B → \mathbb{R}$, there always exists a continuous function $f : \mathbb{R}^2 → \mathbb{R}$ such that $f = g$ on $B$.
(b) Given a continuous function $g : D →\mathbb{R}$, there always exists a continuous function $f : \mathbb{R}^2 → \mathbb{R}$ such that $f = g$ on $D$.
(c) There exists a continous function $f : \mathbb{R}^2 → \mathbb{R}$ such that $f ≡1$ on the set {$(x, y) ∈ \mathbb{R}^2 : x^2+y^2 = 3/2$}
and $f ≡ 0$ on the set
$B$∪{$(x, y) ∈ \mathbb{R}^2 : x^2+y^2≥2$}.


i am totally stuck on this problem.please help somebody.

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What have you tried? Do you have any guesses for each statement? – proximal Jan 7 at 19:16
Perhaps start with a similar question in one dimensional case to build examples. – Maesumi Jan 7 at 22:59
I have no idea how to approach or which theoremmight be helpful – ranadip ganguly Jan 8 at 1:36

1 Answer

For($a$), consider the Tietze Extension Theorem (and this also clues you in for ($b$). For ($c$), consider Urysohn's Lemma.

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I have read these theorem and now try to use them in this problem. for (a) B is a closed subset of the normal space$\mathbb{R}^2$.so by Tietze Extension Theorem it is true. for (b) I can,t apply Tietze Extension Theorem because $D$ is not closed. for (c) Urysohn's Lemma it is true. – ranadip ganguly Jan 8 at 5:32
are my approaches are correct for (a) and (c) – ranadip ganguly Jan 8 at 5:33
and still I am not sure about (b) – ranadip ganguly Jan 8 at 5:33
can anyone help me to prove or disprove (b) – ranadip ganguly Jan 8 at 7:29
You are correct for $a$ and $c$, now all you need to do is find a function where it fails for $b$ (so you want to disprove it). – Clayton Jan 8 at 17:43
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