Open and closed sets
a) ln 0 is undefined, so is the set open for (-infinity,0) union (0, infinity)? I'm a little unsure as when I've put this into an online graphing calculator I get an odd looking graph and the graph actually touches 0. I feel ln x^2 can get arbitrarily close to 0 but never be 0.
b) I want to say this set is open and closed as it covers all the real numbers? It's complement is the empty set which is closed.
Original post by Bameron
a) ln 0 is undefined, so is the set open for (-infinity,0) union (0, infinity)? I'm a little unsure as when I've put this into an online graphing calculator I get an odd looking graph and the graph actually touches 0. I feel ln x^2 can get arbitrarily close to 0 but never be 0.
b) I want to say this set is open and closed as it covers all the real numbers? It's complement is the empty set which is closed.
a) ln 0 is undefined, so is the set open for (-infinity,0) union (0, infinity)? I'm a little unsure as when I've put this into an online graphing calculator I get an odd looking graph and the graph actually touches 0. I feel ln x^2 can get arbitrarily close to 0 but never be 0.
b) I want to say this set is open and closed as it covers all the real numbers? It's complement is the empty set which is closed.
(a) What ln(1)?
(b) Looks correct.
Okay coming back to this...
a) x cannot be -1, 0 or 1, which can be seen as (-infinity,-1)U(-1,0)U(0,1)U(1,+infinity). So any union of open sets is open and Df subset R is open.
b) as said before open and closed as domain is the set (-infinity,+infinity)
c) looking at the first half, ln(9-x^2), the domain is restricted to (-3,0)U(0-3) which is a union of open sets. The second half, sin(x) must be greater than 1/2, looking at [0,2pi], sinx is greater than a half at pi/6 and 5pi/6. To satisfy the second part, x can only be an element of the set [pi/6, 5pi/6]. Bringing this together, it looks like this closed set is the only one that satisfies both the ln part and the sin part of the question, so Df subset R is closed?
a) x cannot be -1, 0 or 1, which can be seen as (-infinity,-1)U(-1,0)U(0,1)U(1,+infinity). So any union of open sets is open and Df subset R is open.
b) as said before open and closed as domain is the set (-infinity,+infinity)
c) looking at the first half, ln(9-x^2), the domain is restricted to (-3,0)U(0-3) which is a union of open sets. The second half, sin(x) must be greater than 1/2, looking at [0,2pi], sinx is greater than a half at pi/6 and 5pi/6. To satisfy the second part, x can only be an element of the set [pi/6, 5pi/6]. Bringing this together, it looks like this closed set is the only one that satisfies both the ln part and the sin part of the question, so Df subset R is closed?
Original post by Bameron
Okay coming back to this...
a) x cannot be -1, 0 or 1, which can be seen as (-infinity,-1)U(-1,0)U(0,1)U(1,+infinity). So any union of open sets is open and Df subset R is open.
b) as said before open and closed as domain is the set (-infinity,+infinity)
c) looking at the first half, ln(9-x^2), the domain is restricted to (-3,0)U(0-3) which is a union of open sets. The second half, sin(x) must be greater than 1/2, looking at [0,2pi], sinx is greater than a half at pi/6 and 5pi/6. To satisfy the second part, x can only be an element of the set [pi/6, 5pi/6]. Bringing this together, it looks like this closed set is the only one that satisfies both the ln part and the sin part of the question, so Df subset R is closed?
a) x cannot be -1, 0 or 1, which can be seen as (-infinity,-1)U(-1,0)U(0,1)U(1,+infinity). So any union of open sets is open and Df subset R is open.
b) as said before open and closed as domain is the set (-infinity,+infinity)
c) looking at the first half, ln(9-x^2), the domain is restricted to (-3,0)U(0-3) which is a union of open sets. The second half, sin(x) must be greater than 1/2, looking at [0,2pi], sinx is greater than a half at pi/6 and 5pi/6. To satisfy the second part, x can only be an element of the set [pi/6, 5pi/6]. Bringing this together, it looks like this closed set is the only one that satisfies both the ln part and the sin part of the question, so Df subset R is closed?
c) First half: what's wrong with taking zero? Otherwise it looks good.
Original post by Bameron
Okay coming back to this...
a) x cannot be -1, 0 or 1, which can be seen as (-infinity,-1)U(-1,0)U(0,1)U(1,+infinity). So any union of open sets is open and Df subset R is open.
a) x cannot be -1, 0 or 1, which can be seen as (-infinity,-1)U(-1,0)U(0,1)U(1,+infinity). So any union of open sets is open and Df subset R is open.
b) as said before open and closed as domain is the set (-infinity,+infinity)
c) looking at the first half, ln(9-x^2), the domain is restricted to (-3,0)U(0-3) which is a union of open sets. The second half, sin(x) must be greater than 1/2, looking at [0,2pi], sinx is greater than a half at pi/6 and 5pi/6. To satisfy the second part, x can only be an element of the set [pi/6, 5pi/6]. Bringing this together, it looks like this closed set is the only one that satisfies both the ln part and the sin part of the question, so Df subset R is closed?The "x can equal 0" issue has been raised.
As far as sin(x) goes, note that it can also equal 1/2; it's kind of clear you realised this when you wrote the closed interval, but it's not what you actually said earlier on.
Note also that in fact (looking only at sin), x can be in any interval of the form [pi/6 + 2npi, 5pi/6 + 2npi].
It does actually turn out here that the only interval that coincides with the " constraint is the one you give, so your final answer is correct. But it's not clear that you properly worked this out (particularly when you say "looking at [0, 2pi]", which definitely leaves some concern that there might be valid values in the range (-3, 0) that you're not considering).
Original post by DFranklin
Something you touched upon when talking about the graph; you could define f(0) = 0, and the resulting function would be continuous at 0. If you did do this, you'd have a function defined at 0 and so the only "forbidden" points would be -1 and 1. There are circumstances in which "defining away the problem at x =0" might make sense, but I don't think this is one of them. In either event, you've still got a union of open sets, so the answer remains the same.
b) as said before open and closed as domain is the set (-infinity,+infinity)
c) looking at the first half, ln(9-x^2), the domain is restricted to (-3,0)U(0-3) which is a union of open sets. The second half, sin(x) must be greater than 1/2, looking at [0,2pi], sinx is greater than a half at pi/6 and 5pi/6. To satisfy the second part, x can only be an element of the set [pi/6, 5pi/6]. Bringing this together, it looks like this closed set is the only one that satisfies both the ln part and the sin part of the question, so Df subset R is closed?
b) as said before open and closed as domain is the set (-infinity,+infinity)
c) looking at the first half, ln(9-x^2), the domain is restricted to (-3,0)U(0-3) which is a union of open sets. The second half, sin(x) must be greater than 1/2, looking at [0,2pi], sinx is greater than a half at pi/6 and 5pi/6. To satisfy the second part, x can only be an element of the set [pi/6, 5pi/6]. Bringing this together, it looks like this closed set is the only one that satisfies both the ln part and the sin part of the question, so Df subset R is closed?
The "x can equal 0" issue has been raised.
As far as sin(x) goes, note that it can also equal 1/2; it's kind of clear you realised this when you wrote the closed interval, but it's not what you actually said earlier on.
Note also that in fact (looking only at sin), x can be in any interval of the form [pi/6 + 2npi, 5pi/6 + 2npi].
It does actually turn out here that the only interval that coincides with the " constraint is the one you give, so your final answer is correct. But it's not clear that you properly worked this out (particularly when you say "looking at [0, 2pi]", which definitely leaves some concern that there might be valid values in the range (-3, 0) that you're not considering).
I really appreciate the detailed response DFranklin!
Yeah sorry I did look at other values of x for the sin(x) part such as -7pi/6 and 13pi/6 which don't satisfy the ln(x^2), but forgot state this in my explanation, thanks once again for your time!
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