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Let $(X, d_X)$ and $(Y, d_Y)$ be metric spaces.
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Suppose that $E \subseteq X$ and $x \in E \setminus E'$.
Let $f : E \to Y$.
The statement "$f$ is continuous at $x$" is...
- ...always true.
- ...nonsense since it only makes sense for $x \in E'$.
- ...nonsense since it only makes sense for $x \in E' \cap E$.
- ...potentially true or false.
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Suppose that $g : \Z \to \R$ is given by \[g(z) = \begin{cases} 0 & \text{ if } z = 0 \\ \frac1{z^2} & \text{ otherwise.} \end{cases}\]
Then $g$ is continuous...
- ...everywhere.
- ...everywhere except $0$.
- ...nowhere.
- Mu. $g$ is not the type of object for which continuity makes sense, so the question is malformed.
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If $X$ is bounded and $g : X \to Y$ is continuous, then $g(X)$ is bounded.
- True.
- True, provided that $g$ is uniformly continuous.
- False.
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Let $f : X \to Y$.
Suppose $(a_n)_n$, $(b_n)_n$ are sequences in $\R_{\gt0}$ so that $(a_n)_n$ converges to $0$ and if $s, t \in X$ with $d_X(s, t) \lt b_n$ it follows that $d_Y( f(s), f(t) ) \lt a_n$.
Which is the strongest true statement below?
- If $(b_n)_n$ also converges to zero, then $f$ is continuous.
- If $(b_n)_n$ also converges to zero, then $f$ is uniformly continuous.
- $f$ is continuous.
- $f$ is uniformly continuous.