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**jsndacruz** **Setup**:

Let $\displaystyle f:\mathbb{R} \to \mathbb{R}$ be continuous. Also, let $\displaystyle S \subseteq \mathbb{R}$ and $\displaystyle T \subseteq \mathbb{R}. $

**1.** If $\displaystyle S$ is open, then $\displaystyle f(S)$ is open.

False: If $\displaystyle f(x) = c $ for some constant $\displaystyle c$, then its range is closed no matter the pre-image.

**2.** If $\displaystyle S$ is closed, then $\displaystyle f(S)$ is closed.

False: Using $\displaystyle f(x)=e^x$ and $\displaystyle S = ( - {\infty} , 0 )$, then $\displaystyle f(S) = (0,1)$.

**3.** If $\displaystyle S$ is bounded, then $\displaystyle f(S)$ is bounded.

False: __But__ I'm having difficulty thinking of an example.

**4.** If $\displaystyle T$ is open, then $\displaystyle f^-1 (T)$ is open.

True: __But__ again, I'm not sure where to turn for the proof.

**5.** If $\displaystyle T$ is closed, then $\displaystyle f^-1 (T)$ is closed.

True: If I had the above, I'd use its complement to show this one holds, but since I don't... Let $\displaystyle x \in f^-1 (T)$ and let $\displaystyle x_n$ be a sequence of elements from $\displaystyle x \in f^-1 (T)$ that converges to $\displaystyle x$. Since $\displaystyle f$ is continuous, $\displaystyle f(x_==nn)$ converges to $\displaystyle f(x)$. Since $\displaystyle C$ is closed, $\displaystyle f(x)$ is in C. So, $\displaystyle x$ is in $\displaystyle f^-1 (C) $.