choose the correct option of given function. $A = f(B) \subseteq \mathbb R$ where $B$ is a closed interval contained in $(0, \infty)$ and $f(t)= \log t$.


*

*Open

*closed 

*connected 

*compact


From my point of view $f(t)$ must be closed, by the theorem continuous image of a closed set is closed , closed map  to closed map , but I don't know  whether it is compact or connected . I have no idea any other idea about this compact and connected. I know that a circle and an ellipse are  both compact and connected. Here this is not mentioned, so I'm very confused.
If anbody help me I would be very thankful to him.
 A: Ok we have that if $a>0$ then $B=[a, b]$ which is a closed interval.
Also $b$ can be $\infty$.
The image $f(B)$ is not always compact under $f(t)=\log{t}$
Take for instance $B=[1,+ \infty)$ which is  closed.But $f(B)=[0,+ \infty)$  is not compact because from  Heine-Borel theorem  ,a subset of the real line is compact iff is closed and bounded.Thus we cannot choose $4$.It is not always the case.
Now from a theorem in general topology we know that a continuous image of a connected set,is also connected.
Every interval $B$ in $\mathbb{R}$(in the case every subset of $(0, +\infty)$)   is connected thus  $f(B)$ is connected.So $3$ is correct.
Because $f$ is invertible and continuous you can prove that $f(B)$ is closed if $B$ is closed.
But if $B$ is closed then it cannot be open because $\mathbb{R}$ is connected thus from definition of connectedness in a topological space we have that a space is connected if the only clopen(closed and open at the same time) subsets of the space are the space itself and the empty set.
Thus $1$ is not the case.
I hope it helps.
