a .$ f (U) $ is open

b. $f (U) $ is closed

c. $f^-1(U)$ is open up

d. $f^-1(U)$ is closed

Attempt is that I take any set $U$ consisting of facets say $1,2,3$ then $f (U)$ will have actually $1 ,4,9$ .Now limit suggest of $f (U)$ is $phi $ and also hence it is closed but answer appears to be $c$.

You watching: How to determine if a set is open or closed

Can anyone aid where I went wrong?

Hint: If $f$ is consistent then $f^-1(U)$ is open to any type of $U subset sarkariresultonline.infobbR$ open up set.

**Edit:** Consider $f: M o N$ consistent.

Let $A" subset N$ be an open up collection, we desire to present that $f^-1(A")$ is open. In fact, for each $a in f^-1(A")$, we have that $f(a) in A"$. By meaning of open set, there exists $epsilon > 0$ such that $B(f(a), epsilon) subcollection A"$. As $f$ is constant at $a$, there exists $delta > 0$ correspondent to $epsilon > 0$ such that

$$f(B(a; delta)) subset B(f(a); epsilon) subcollection A"$$

that is, $$B(a;delta) subcollection f^-1(A")$$

Then $f^-1(A")$ is open up.

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edited Dec 24 "14 at 17:13

answered Dec 24 "14 at 16:44

Aaron MarojaAaron Maroja

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Statement c is of course true, because the attribute is consistent.

If you understand that the attribute is consistent (at eextremely point), then take an open collection $U$. In order to present that $f^-1(U)$ is open, we should check out that, for $xin f^-1(U)$, tbelow exists $delta>0$ with $$B(x,delta)={rinsarkariresultonline.infobbR:|r-x|0$ such that$$B(f(x),varepsilon)subseteq U.$$Now apply the $varepsilon$-$delta$ definition of continuity and the problem is solved: the required $delta>0$ is specifically the one provided by the continuity problem.

Why is the attribute continuous? By general theorems: the product of constant functions is continuous; the identification attribute $tmapsto t$ is obviously constant.

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The other statements are false, in the sense that for each one you have the right to discover a specific $U$ such that the condition does not organize.

For a think about $U=sarkariresultonline.infobbR$. Hint: $f(U)=<0,infty)$

For b take into consideration $U=(0,infty)$. Hint: $f(U)=(0,infty)$

For d think about $U=(0,1)$. Hint: $f^-1(U)=(-1,0)cup(0,1)$

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edited Dec 24 "14 at 18:43

answered Dec 24 "14 at 17:45

egregegreg

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