Why is zero not included in the interval (0, infinity)?

[Casio1]

g(x) = sqrt x

Because x in the rule is square root on the right hand side I can understand why only a positive number would be allowed in the left hand side value of x.

I undertand that because x cannot be less than 0, but the square root of x can be any positive number, then the interval notation I understand would be [0, infinity)

zero because x cannot be less than 0, and infinity because x can be any other number.

OK, this is where I have a problem understanding.

h(x) = 1/x

x cannot be zero, the function h has domain R excluding 0, I am to understand this consists of two open intervals.

I understand that x can be any negative number, but am failing to understand why 0 is then included, the interval is written as ( - infinity, 0)

I am told that R excludes 0, so then how can the open interval include (0, infinity).

I understand the infinity part, but x cannot be zero so where does that arrive in the interval?

Please advise if you can.

 

[Plato2]

g(x) = sqrt x
Because x in the rule is square root on the right hand side I can understand why only a positive number would be allowed in the left hand side value of x.
I undertand that because x cannot be less than 0, but the square root of x can be any positive number, then the interval notation I understand would be [0, infinity)
zero because x cannot be less than 0, and infinity because x can be any other number.
OK, this is where I have a problem understanding.
h(x) = 1/x
x cannot be zero, the function h has domain R excluding 0, I am to understand this consists of two open intervals.
I understand that x can be any negative number, but am failing to understand why 0 is then included, the interval is written as ( - infinity, 0)
I am told that R excludes 0, so then how can the open interval include (0, infinity).
I understand the infinity part, but x cannot be zero so where does that arrive in the interval?
Please advise if you can.

$(-\infty)\cup(0,\infty)$

 

[Opalg]

g(x) = sqrt x

Because x in the rule is square root on the right hand side I can understand why only a positive number would be allowed in the left hand side value of x.

I undertand that because x cannot be less than 0, but the square root of x can be any positive number, then the interval notation I understand would be [0, infinity)

zero because x cannot be less than 0, and infinity because x can be any other number.

OK, this is where I have a problem understanding.

h(x) = 1/x

x cannot be zero, the function h has domain R excluding 0, I am to understand this consists of two open intervals.

I understand that x can be any negative number, but am failing to understand why 0 is then included, the interval is written as ( - infinity, 0)

I am told that R excludes 0, so then how can the open interval include (0, infinity).

I understand the infinity part, but x cannot be zero so where does that arrive in the interval?

Please advise if you can.

I think that you are maybe failing to notice the difference between [0, infinity) and (0, infinity). A square bracket means that the endpoint is included in the interval (as in a closed interval), but a parenthesis means that the endpoint is excluded (as in an open interval). To be precise, (0, infinity) means the set of all numbers x such that 0<x<infinity. The strict inequality means that 0 is not included in the interval.

 

[Casio1]

I think that you are maybe failing to notice the difference between [0, infinity) and (0, infinity). A square bracket means that the endpoint is included in the interval (as in a closed interval), but a parenthesis means that the endpoint is excluded (as in an open interval). To be precise, (0, infinity) means the set of all numbers x such that 0<x<infinity. The strict inequality means that 0 is not included in the interval.

I am sorry I have probably not explained the misunderstanding too well

h(x) = 1/x

if h(0) = 1/0, then this would not be allowed, thus anything divided by zero = 0

I am told that the function h has domain R excluding 0, which consists of two open intervals (- infinity, 0) and (0, infinity)

I understand the parenthesis, they are open intervals, and I understand the - infinity as the x value can be any negative number up to 0.

The confusion is that I am told it excludes 0 yet 0 is included in the parenthesis indicating to me that the interval finishes at 0, then starts at 0 again to infinity, yet the question says R excludes 0.

That is what I am confused about?

 

[Evgeny.Makarov]

The confusion is that I am told it excludes 0 yet 0 is included in the parenthesis indicating to me that the interval finishes at 0, then starts at 0 again to infinity, yet the question says R excludes 0.

What exactly do you mean by "0 is included in the parenthesis" and "the interval finishes at 0"?

 

[Reckoner]

I am sorry I have probably not explained the misunderstanding too well

I suggest you read Opalg's reply again, very carefully. I believe he addresses the source of your confusion. The parentheses indicate that we are to include all numbers up to, but not including 0. So no matter how close you get to zero, the number is included in the domain, until you get to zero itself. \((0, \infty)\) would, for example, include numbers like 0.0000001, or \(10^{-50}\), or really any real number strictly greater than zero.

You may then wonder, perhaps, why in our notation we write the interval as starting from 0 instead of starting from the "next" number after zero. The answer is that there is no such number. No matter what positive number you come up with, it is always possible to find a smaller one. So we use zero as the boundary, with the understanding that zero itself is not to be included.