Is the Graph of a Continuous Function Closed in R^2?

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SUMMARY

The graph of a continuous function \( f: \mathbb{R} \to \mathbb{R} \), denoted as \( G = \{(x, f(x)) | x \in \mathbb{R}\} \), is proven to be closed in \( \mathbb{R}^2 \) under the Euclidean metric. This conclusion is derived from the property that continuity preserves limits; specifically, if a sequence \( (x_n) \) converges to \( x \) in \( \mathbb{R} \), then \( f(x_n) \) converges to \( f(x) \). The discussion emphasizes that if the complement \( \mathbb{R}^2 - G \) is open, then \( G \) must be closed, utilizing the characterization of closed sets in metric spaces.

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The graph of a continuous funtions (R -> R) is the subset G:={(x, f(x) | x element of R} is a subset of R^2. Prove that if f is continuous, then G is closed in R^2 (with euclidean metric).


I know that continuity preserves limits, so xn -> x in X means f(xn-> y in Y.
and for all A element of R^2 - G there exists r > 0 st B(a, r) subset of R^2 - G.I know that if R^2 - G is open then G is closed.


What is the connection between limits and the open ball B? I think that might be the way...
 
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the following characterization might be useful:

A subset F of a metric space X is closed iff for all sequences (x_n) in F which converges to a certain x in X, it holds that x is in F.
 

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