continuous maps compact set to compact set.

Question

If $A$ is a subset of $\Bbb R$ and $f : A \to \Bbb R$ is a continuous function then prove or disprove the following.

1. If $A$ is a bounded set but not closed then $f(A)$ is bounded.

2. If $A$ is a closed set but not bounded then $f(A)$ is closed.

My Attempt : I have seen a result which tells that Image of closed and bounded set is closed and bounded. Will happen when one condition either closed or bounded is relaxed.

Answer 1

1) Take $A=(0,\pi/2)$ Then $f(x)=\tan(x)$ is unbounded, but $A$ is bounded.

2) Take $A=\Bbb R$ then $A$ is closed (though also open) and the map $f(x)=\frac{1}{1+e^{-x}}$ has $f(A)=(0,1)$ which is open.

Answer 2

A brief answer to settle one of the questions that Baljeet asks: in general, a continuous function $f: \mathbb{R} \to \mathbb{R}$ will not be closed (i.e. enjoying the property of sending closed subsets to closed subsets). Here is one possible counter-example: consider any $a \in (0,1)$ and take $f$ to be the exponential function of base $a$, in other words such that

$$f: \mathbb{R} \to \mathbb{R} \\ f(x)=a^x$$

Clearly $\mathbb{N} \subset \mathbb{R}$ is a closed subset in the order topology (the standard one at work in analysis), however $f(\mathbb{N})=\{a^n\}_{n \in \mathbb{N}}$ is no longer closed: since $$a^n \xrightarrow{n \to \infty} 0$$

we infer that $0$ is a point adherent to this subset (i.e. in its closure) lying though in the complement of this subset.

A perhaps even clearer counter-example can be found in the instance of $$g: \mathbb{R} \to \mathbb{R} \\ g(x)=\frac{1}{x^2+1}$$

Indeed $g([0, \infty))=(0,1]$ is not closed.