# Problems with showing that the closed graph of bounded $f$ implies continuity of $f$

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.

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 $$y=f(x)$$.

• Thx, but can you teach me why $f(\mathbb{R})$ is complete? – Moreblue Nov 4 '18 at 8:42
• @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. – Alvin Lepik Nov 4 '18 at 8:56
• 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? – Moreblue Nov 4 '18 at 9:10
• @Moreblue I redid the proof using more real-analysis friendly language, I'll let you fill in the details. – Alvin Lepik Nov 4 '18 at 10:11