The question is in the given picture :
And the answer is C as shown in the answer sheet below:
But this answer is so strange for me as I know that a set is closed if and only if it contains all its limit points and so the choice (A) is true also.
Could anyone explain this for me please? I want why should I exclude every choice other than (C).
On
Let's think of a tangible example.
Let $B = (0,1)$, and $b = 1$.
A. Is $B$ closed? Evidently not.
B. Is $B$ not open? In this case no. I could have written that $B' = [0,1)$ though, which is not open. Either case, this doesn't have to be true.
C. Is $b$ a limit point of $B$? It has to be, otherwise it wouldn't be the least upper bound.
D. This isn't true - for our example, consider $b_n = 1-\frac{1}{n}$. This is in $B$, and converges to $b$.
E. This isn't true. We'd want an interval of the form $(1,2)$ or something like that, but $b$ isn't in this. If we try $(x,2)$ for $x<1$, $b$ will be in this, but it will always share another element of $B$.
For questions like (multiple choice), thinking of tangible examples can help a ton.
The question states that $b$ is not in $B$. Since $b$ is a limit point of $B$, $B$ does not contain all its limits points. So $B$ is not closed.
EDIT:
If there is an open interval $(b-\epsilon,b+\epsilon)$ with no element of $B$, then $b-\epsilon$ is an upper bound for $B$, which is a contradiction. So every open interval around $b$ intersects $B$, that is, $b$ is a limit point of $B$. This proves that C is true.
$B$ is not closed, because $b$ is a limit point of $B$ but $b\notin B$. So A is false.
Under the hypotesis of the question, $B$ may or may not be open. For example, $(b-1,b)$ is open, $b$ is its least upper bound and $b\notin (b-1,b)$. So, although we can not say that B is false, we can not either say that it is true. So B is not the correct answer.
D is false. We can define $b_n$ to be in the intersection $(b-\frac1n,b)\cap B$ to build a sequence in $B$ that converges to $b$.
E is also false, as we have already seen.