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I know $f:[1,5]\to\mathbb R$, means $f$ is a function from $[1,5]$ to $\mathbb R$. I am just abit unclear now on the exact interpretation of "to $\mathbb R$". Is $1\le x\le 5$ the domain? And is $\mathbb R$ the co-domain (or image?)?

Is my interpretation in words ---$f$ is a function which takes a number $1\le x\le 5$, and maps it onto the real numbers $\mathbb R$, correct?

Suppose we take $f=x^2$ and $x=2$, does $f:[1,5]\to\mathbb R$ hold? So thus the function gives us $4$, which is $\in \mathbb R$.

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    Should we combine this question with http://math.stackexchange.com/questions/95606/meaning-of-fa-b-to-mathbbr2012-01-01

4 Answers 4

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Careful with "codomain" and "image". Codomain always means a set that contains all values of the function; "image" may mean only the elements that are actually outputs of the function, or it may mean the same as codomain.

"$f\colon [1,5]\to\mathbb{R}$" means that $f$ is

  1. a function;
  2. the domain of $f$ is the whole set $[1,5]$; and
  3. the value of $f$ at each element of the domain is some element of $\mathbb{R}$.

So you can plug into $f$ any number $a$, $1\leq a\leq 5$, and those are the only things you can plug into $f$. And when you plug such a number into $f$, what "comes out" will be a real number.

If $f$ is given by the rule "if $x$ is in $[1,5]$, then $f(x)$ is $x^2$", then $f$ is indeed a function with domain $[1,5]$ and images contained in $\mathbb{R}$, so you would be justified in writing $f\colon[1,5]\to \mathbb{R}$.

Writing "$f=x^2$ and $x=2$" is at best confusing. It's better to write $f(x)=x^2$ (indicating that $f$ is a function with input $x$). Much better is to specify the domain, since a standard convention is that when you write down a formula such as $f(x)=x^2$, the domain is understood to be "all real numbers for which the formula makes sense and gives a real number as an output" (this is very common in real analysis and calculus, for example), so simply writing "$f(x)=x^2$" would immediately imply a function whose domain is all real numbers. To specify domain, you would write something like $f(x) = x^2,\qquad 1\leq x\leq 5,$ or $f(x) = x^2,\qquad x\in [1,5].$

If what you mean to write was something like "$f(x)=x^2$ and $x=2$", then you are describing a function whose domain is $\{2\}$ and not $[1,5]$; so you would not be justified in claiming this function can be described as a function $f\colon[1,5]\to\mathbb{R}$.

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The notation $ f: [1,5] \rightarrow \mathbb{R} $ means that $f$ is a function whose domain is to taken to be the interval $[1,5]$ and whose codomain is $\mathbb{R}$ (i.e. all the outputs of $f$ fall into $\mathbb{R}$). It makes no claims about surjectivity or injectivity; you must analyze the function itself to decide that.

To address your example, you define $ f: [1,5] \rightarrow ? $ $ f(x) = x^2, $ where the question mark means we aren't sure what to put there yet. Since since the square of any number in the interval $[1,5]$ is a real number, then indeed $\mathbb{R}$ is an acceptable codomain for $f$, so we could write $ f: [1,5] \rightarrow \mathbb{R}. $ Notice that, as we have defined it, $f$ is not surjective, since some numbers in $\mathbb{R}$ cannot be reached by $f$ (like 36).

Looking a little closer, we might notice that the square of any number in the interval $[1,5]$ falls in the interval $[1,25]$. Thus, we would also be correct in writing $ f: [1,5] \rightarrow [1,25]. $ After this change in codomain, $f$ becomes surjective, since every number in $[1,25]$ can be reached by $f$.

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It's correct if you say "into" rather than "onto". The domain is $[1,5]=\{x: 1\le x\le 5\}$. For every number $x$ in the domain, $f(x)$ is in $\mathbb{R}$.

"Onto" is often taken to mean "surjective" (except that "onto" is also used as a preposition). I.e. if $f$ is said to map the set $[0,5]$ "into" $\mathbb{R}$, that means only that for every number $x$ in the domain, $f(x)$ is in $\mathbb{R}$, but if $f$ is said to map the set $[0,5]$ "onto" $\mathbb{R}$, that means that in addition to the fact that for every number $x$ in the domain, $f(x)$ is in $\mathbb{R}$, it is also the case that for every member $y$ of $\mathbb{R}$, there is some number $x$ in $[1,5]$ such that $f(x)=y$. So, for example $f(x)=x^2$ maps $[0,1]$ into, but not onto, $\mathbb{R}$. It's not onto because for example $26\in\mathbb{R}$ but there is no number $x\in[1,5]$ such that $f(x)=26$.

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Your function $f$ can be defined on all sorts of domains. You can define it on [2,2] (i.e. at the point 2) or on the closed interval [1,5] (real numbers from 1 to 5 including 1 and 5). You can define it on subsets of the Real or Complex numbers. But if you are worried about such things with such functions you will give yourself a headache.

You could also (for completeness) define it on an interval like (1,5) - real numbers between 1 and 5 but not including 1 or 5.

Whichever domain of definition you were to choose, the vale at the Real number 2 would be the same (or at 2 identified as a complex number on the Real axis).

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    @The Chas: Thanks.2012-01-01