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

False. The characteristic function serves as a counterexample
$\displaystyle f(x)= \begin{cases} 1 & \text{if$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 $\displaystyle f:\mathbb{R} \rightarrow \mathbb{R}$ f(A) is closed whenever A is closed, then f is continuous.
False. Consider the function
$\displaystyle f(x)= \begin{cases} tanx & \text{if$ - \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 $\displaystyle f:\mathbb{R} \rightarrow \mathbb{R}$ f(A) is open whenever A is open, then f is continuous.
False. I was thinking f(x) = 1/x. The range is $\displaystyle (- \infty, 0) \cup (0, \infty)$, which is open, but f is not continuous.

Thank you.

2. Originally Posted by Paperwings
1. If $\displaystyle f:\mathbb{R} \rightarrow \mathbb{R}$ 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 $\displaystyle 1$ for rationals and $\displaystyle 0$ for irrationals. It is nowhere continous.

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

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

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