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Math Help - True/False - continuous function

  1. #1
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    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)=<br />
\begin{cases}<br />
1 & \text{if $x \in K$}, \\<br />
0 & \text{if $x \notin K$}.<br />
\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)=<br />
\begin{cases}<br />
tanx & \text{if $ - \frac{\pi}{2} < x < \frac{\pi}{2}$}, \\<br />
0 & \text{if $otherwise$}.<br />
\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.
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  2. #2
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    Quote Originally Posted by Paperwings View Post
    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.
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