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Let, $A=f(B)\subset \mathbb R$ ,where $B$ is closed interval contained in $(0,\infty)$ and $f(t)=\log t$. Then $A$ is

(a) closed

(b) open

(c) compact

(d) connected ?

My Attempt:

Clearly $f$ is continuous. Suppose that $B=[a,b]\subset(0,\infty)$. Then $B$ is connected & so $A=f(B)$ is connected.

As any closed interval in $\mathbb R$ is closed & bounded so, compact & hence $A$ is compact. Continuous image of open or closed set is not necessarily open or closed.

So options (c) & (d) are correct. But the answer is given (a) & (d).

Where my mistake?

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    $\begingroup$ $[3,\infty)$ is closed but not compact.we need to show that for all closed subsets $\endgroup$ – BigM Feb 13 '15 at 16:04
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Any closed interval in $\mathbb{R}$ is not closed and bounded -- Example: $[1, \infty)$ is closed, but it is not bounded.

The compact sets in $\mathbb{R}^n$ are those which are closed and bounded.

(d) is correct since continuous image of connected set is connected and intervals are connected.

Note that if $B=[a,b], A=[\log a, \log b]$,and $B=(0,a]$ then $A=(-\infty, \log a]$ and $B=[a, \infty)$ then $A=[\log a, \infty)$. So, the image of $B$ under $f$ is a closed interval so (a) is true. (b) is not true, since the only closed and open sets in $\mathbb{R}$ are $\mathbb{R}$ and $\emptyset$. Clearly, $[a,\infty)$ is not compact (it isn't bounded), so (c) is wrong.

So, the correct answers are (a) and (d).

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