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I have the following problem

Given bounded $f : \mathbb{R} \to \mathbb{R}$, show : $\{ (x, f(x)) \in \mathbb{R}^2 : x\in\mathbb{R} \}$ : closed then $f$ : continuous on $\mathbb{R}$.


Try

Consider $\langle x_n \rangle \subset \mathbb{R}$ s.t. $\lim_{n \to \infty} x_n = x_0$

Since $\{f(x) : x \in \mathbb{R} \}$ : compact, thus there exists $r : \mathbb{N} \to \mathbb{N}$, strictly increasing s.t. $f(x_{r(n)}) \to y_0 \in \{f(x) : x \in \mathbb{R} \}$.

Let $z_n := (x_n, f(x_n))$ then $z_{r(n)} \to (x_0, y_0)$

But I cannot see how I should proceed.

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Pick $x_n\xrightarrow[n\to\infty]{}x$ (in the domain). We show $f(x_n)\xrightarrow[n\to\infty]{}f(x)$.
Since $f(\mathbb R)$ is bounded the sequence $(x_n,f(x_n))$ is bounded, thus we may assume (per Bolzano-Weierstrass) that $$(x_n,f(x_n))\xrightarrow[n\to\infty]{} (x,y). $$ Since the graph is closed we have $(x,y)\in\text{gr}f$. Since limits are unique we have $y=f(x)$.

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  • $\begingroup$ Thx, but can you teach me why $f(\mathbb{R})$ is complete? $\endgroup$ – Moreblue Nov 4 '18 at 8:42
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    $\begingroup$ @Moreblue I misspoke. We have that the image is bounded per assumption. Also it is closed because the graph is closed. Thus it is a closed subset in a complete space, hence complete. It is also bounded, therefore compact. $\endgroup$ – Alvin Lepik Nov 4 '18 at 8:56
  • $\begingroup$ But what I'm stuck at is why $f(\mathbb{R})$ is closed if graph is closed, even though I somehow groundlessly mentioned $f(X)$ is compact. Would you specify it a little more? $\endgroup$ – Moreblue Nov 4 '18 at 9:10
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    $\begingroup$ @Moreblue I redid the proof using more real-analysis friendly language, I'll let you fill in the details. $\endgroup$ – Alvin Lepik Nov 4 '18 at 10:11

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