TITLE: Can endpoints be local minimum?
QUESTION [8 upvotes]: My textbook defines local maximum as follows:

A function $f$ has local maximum value at point $c$ within its
  domain $D$ if $f(x)\leq f(c)$ for all $x$ in its domain lying in some
  open interval containing $c$.

The question asks to find any local maximum or minimum values in the function 
$$g(x)=x^2-4x+4$$ in the domain $1\leq x<+\infty$.
The answer at the back has the point $(1,1)$, which is the endpoint. 
According to the definition given in the textbook, I would think endpoints cannot be local minimum or maximum given that they cannot be in an open interval containing themselves. (ex: the open interval $(1,3)$ does not contain $1$). Where am I wrong?

REPLY [3 votes]: I think fundamentally the comments are right, and you should speak with your teacher to confirm definitions and expectations.  But there's also a point to make about topology here, which could justify the book's definition and answer as consistent.  
The definition of local maximum you gave is:

A function $f$ has a local maximum at point $c$ within its domain $D$ if $f(x) \leq f(c)$ for all $x$ in its domain lying in some ** open ** interval containing $c$.  

If you interpret this as saying that the interval can come from $\mathbb{R}$, and is not restricted to $D$, then you have no problem, as others have pointed out.  But like you I am thinking about being restricted to $D$ and my instinct is to think only about intervals in $D$.  This can still be ok, if we just alter our interpretation of "open" a little bit (in a natural way)...
Now, whenever we say "open" we're really saying "open with respect to ** insert topology here ** ."  A lot of the time it's obvious from context or the textbook has established a practice of contextual implication, but in this case (without knowing your book) I'd argue there are two reasonable interpretations:

We might be talking open intervals with respect to the standard topology on $\mathbb{R}$ (which is what you've probably been using in your class), but 
since we're restricting our attention to a domain $D \subset \mathbb{R}$, it's also pretty normal to talk about a different topology, called the subset topology on $D$ (induced by the standard topology on $R$).  


In the subset topology on $D \subset \mathbb{R}$ (induced by the standard topology), a set $S$ is open if and only if $S$ is the intersection $D \cap X $, with $X$ open in $\mathbb{R}$ with respect to the standard topology on $\mathbb{R}$.  

We're often more interested in the subset topology than the usual topology on the whole space just because of situations like the one you're in, in which a definition doesn't work quite like you expect when $D \not= \mathbb{R}$.  
So let's work with a slightly different definition of local maximum:

A function $f$ has a local maximum at point $c$ within its domain $D$ if $f(x) \leq f(c)$ for all $x$ in its domain lying in some interval $I$ containing $c$ such that $I$ is open with respect to the subset topology on $D$.

Now back to your case.  Let $D = [1, \infty)$.  For any $a > 1$, we have that 
$$[1,a) = D \cap (-a,a)$$
 Since $(-a,a)$ is open in $\mathbb{R}$ with respect to the standard topology, $[1,a)$ is open in $D$ with respect to the subset topology on $D$. This intuitively makes sense, because if you were an ant walking on $f(D)$, when you came to $f(1)$ you'd have nowhere to go but down.