# Thread: True/False - continuous function

1. ## True/False - continuous function

The following are true or false variances of determining whether a function is continuous.

1. If $f:\mathbb{R} \rightarrow \mathbb{R}$ f(K) is bounded, then f is continuous.
False. A bounded step function such as the characteristic function would serve as a counterexample.

2. If $f:\mathbb{R} \rightarrow \mathbb{R}$ f(K) is sequentially compact whenever K is sequentially, then f is continuous.

False. The characteristic function serves as a counterexample
$f(x)=
\begin{cases}
1 & \text{if x \in K}, \\
0 & \text{if x \notin K}.
\end{cases}$
since it is bounded and closed, but it is not continuous.

(I know the converse is true: if f is continuous and K is sequentially compact, then f(K) is sequentially compact. )

3. If $f:\mathbb{R} \rightarrow \mathbb{R}$ f(A) is closed whenever A is closed, then f is continuous.
False. Consider the function
$f(x)=
\begin{cases}
tanx & \text{if - \frac{\pi}{2} < x < \frac{\pi}{2}}, \\
0 & \text{if otherwise}.
\end{cases}$

It is closed on the real line but it is not continuous. Is there a better counterexample for this?

4. If $f:\mathbb{R} \rightarrow \mathbb{R}$ f(A) is open whenever A is open, then f is continuous.
False. I was thinking f(x) = 1/x. The range is $(- \infty, 0) \cup (0, \infty)$, which is open, but f is not continuous.

Thank you.

2. Originally Posted by Paperwings
1. If $f:\mathbb{R} \rightarrow \mathbb{R}$ f(K) is bounded, then f is continuous.
False. A bounded step function such as the characteristic function would serve as a counterexample.
I would use the Dirichlet function. It is defined to be $1$ for rationals and $0$ for irrationals. It is nowhere continous.

2. If $f:\mathbb{R} \rightarrow \mathbb{R}$ f(K) is sequentially compact whenever K is sequentially, then f is continuous.
Again use Dirichlet.

3. If $f:\mathbb{R} \rightarrow \mathbb{R}$ f(A) is closed whenever A is closed, then f is continuous.
Again use Dirichlet.

4. If $f:\mathbb{R} \rightarrow \mathbb{R}$ f(A) is open whenever A is open, then f is continuous.
Define $f(x) = 1/x$ and $f(0)=0$.
Then $f(\mathbb{R}) = \mathbb{R}$.
And it is clearly not continous.