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Which properties hold for the following set?
Open, connected, compact, closed
$A=f(B)⊂\mathbb{R}$ where $B$ is a closed interval contained in $(0,∞)$ and $f(t)=\log t$

$\log t$ is a continuous for a closed interval $A$ must be closed.
$\log t$ is not bounded on $(0,∞)$. so it is not compact.
it is not open in closed bounded interval. since every interval is connected so $A$ is connected.

so $A$ is only connected and closed.

am I correct?

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You write that $\log(t)$ is not bounded on $(0,\infty )$, but your question is whether $\log(t)$ is bounded on $B$. – Todd Wilcox Jan 2 '13 at 15:05
up vote 3 down vote accepted

$f(t) = \log(t)$ is a homeomorphism between a subset $B$ of $(0,\infty)$ and its image $A = f[B]$. This holds in general (its inverse is $e^t$).

So $A$ has the same topological properties as $B$ has. As $B$ is a closed interval, it is compact (supposing that a set like $[1, \infty)$ is not called an interval but a (closed) segment). Any interval is connected. Compact implies closed (in Hausdorff spaces like $\mathbb{R}$) so $B$ (and also $A$) is closed. It cannot be open, by connectedness of $\mathbb{R}$ and the image is compact, so not $\mathbb{R}$.

So closed, compact, connected.

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why compact? if I take the interval[$1,\infty$) then it is a closed interval contained in ($0,∞$) but not compact. am I right or missing something? – poton Jan 2 '13 at 15:12
Like I said, I don't call that an interval, but maybe your text does. Then closed and connected is the only thing we can say, and compactness goes away. – Henno Brandsma Jan 2 '13 at 15:32

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