If A is closed, if f(A) necessarily closed?

Question

Let $A \subseteq \mathbb{R}$. Let $f:A \rightarrow \mathbb{R}$ be continuous function.

If $A$ is bounded, is $f(A)$ necessarily bounded?

If $A$ is closed, is $f(A)$ necessarily closed?

Does $f$ must be uniformly continuous to preserve those properties? and not only continuous? I manage to figure out that if $f$ uniformly continuous,then $f(A)$ is bounded.

However, I am stuck on the first part of the question.

Answer 1

HINT: Try an unbounded closed subset for $A$.

Answer 2

Let $f(x)=e^{-1/(x^{2}+1)}$ for $x\in\mathbb{R}$, $1$ is an accumulation point of $f(\mathbb{R})$ which does not belong to $f(\mathbb{R})$, so $f(\mathbb{R})$ is not closed.

Answer 3

$(0,1)$ is bounded and $f: (0,1)\to \mathbb{R}: x \mapsto \frac{1}{x}$ is unbounded.

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$\mathbb{R}$ is closed and $\arctan: \mathbb{R} \to [\frac{-\pi}{2},\frac{\pi}{2}]$ maps $\mathbb{R}$ onto the set $(\frac{-\pi}{2},\frac{\pi}{2})$, which is not closed.

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Interestingly, if $A \subseteq \mathbb{R}$ is closed and bounded, it is compact, thus its image $f(A)$ is also compact by continuity, so $f(A)$ is closed and bounded in $\mathbb{R}$.

I guess this was the point of the exercise you were given.