Any help with the following problem:
Prove that if a function $f: \mathbb{R}\rightarrow \mathbb{R}$ is bounded and its graph is a closed subset of $\mathbb{R}^{2}$, then $f$ is continuous.
Any help with the following problem:
Prove that if a function $f: \mathbb{R}\rightarrow \mathbb{R}$ is bounded and its graph is a closed subset of $\mathbb{R}^{2}$, then $f$ is continuous.
HINT: Try proving the contrapositive, namely
If $f: \mathbb{R} \rightarrow \mathbb{R}$ has a point of discontinuity, then $f$ is either unbounded OR its graph is not a closed subset of $\mathbb{R}^2$.
A common way to deal with "OR" in a conclusion is to break the proof into two cases. First, suppose $f$ is unbounded. Since this is one of the two conclusions we hope to draw, we're already done! In the other case, we can assume $f$ is bounded and hope to conclude that $f$ is not a closed subset of $\mathbb{R}^2$. In other words, we can transform the original problem into this one:
If $f: \mathbb{R} \rightarrow \mathbb{R}$ has a point of discontinuity AND $f$ is bounded, then $f$ is not a closed subset of $\mathbb{R}^2$.
There is a useful theorem due to Kuratowski: if $C$ is compact and $X$ is any space, then the projection $\pi: C \times X \rightarrow X$ defined by $\pi(x,y) = y$ is a closed (and continuous, of course) function. See here for a short proof.
If $f$ is bounded let $C$ be a compact interval with $f[X] \subset C$. The graph $\Gamma =\{(x, f(x)) \mid x \in \mathbb{R} \}$ of $f$ is a subset of $\mathbb{R} \times C$ by assumption. If now $K$ is a closed subset of $C$, then note that we have
$f^{-1}[K] = \pi[ \Gamma \cap (\mathbb{R} \times K) ]$
(as $x \in f^{-1}[K]$ then $f(x) \in K$ and $(x, f(x))$ is in $\Gamma$ and in $\mathbb{R} \times K$ and $x$ is its image under $\pi$, while if $x = \pi(x,y)$ for $y \in \Gamma \cap \mathbb{R} \times K$ we know $y \in K$ and $y = f(x)$ to be on $\Gamma$, so $x \in f^{-1}[K]$)
Also, as $\pi$ is a closed map and the set we take an image of is closed in $\mathbb{R} \times C$, $f^{-1}[K]$ is closed in $\mathbb{R}$ for all closed subsets $K$ of $C$, so $f$ is a continuous map from $\mathbb{R}$ to $C$ and thus to $\mathbb{R}$ as well.
Note that the fact that we use $\mathbb{R}$ is irrelevant, all we need is that the image of $f$ is contained in a compact set in the codomain, and there are no conditions on the domain. The condition is needed because of the example $f(x) = \frac{1}{x}$ for $ x\neq 0$ and $f(0)= 0$ which has a closed graph but is not continuous, and $f$ is not bounded.
Interesting question; I haven't seen it before, so here's my stab at it.
It's clear that the boundedness is needed, because otherwise we have the counterexample:
$f(x) = \begin{cases} 0 & \text{if } x = n\pi \\ \tan(x) & \text{otherwise} \end{cases}$
So we're given that $f$ is bounded and its graph, which I'll call $G$, is a closed subset of $\mathbb{R}^2$.
In order to show that it's continuous, we can show any one of a list of equivalent conditions, including the inverse image of an open set is open, or the inverse image of a closed set is closed. But $f^{-1}(S)$, for $S \subset \mathbb{R}$, is just the projection of $(\mathbb{R} \times S) \cap G$ onto the $x$-axis.
If you utilize the fact that $G \subset \mathbb{R} \times I$, where $I$ is a sufficiently large closed interval, and consider $f^{-1}(C)$, $C$ a closed subset of $I$, I believe compactness is on your side to show that $f^{-1}(C)$ contains all of its limit points. See if you can work out the details from there.
Alternative to the answers already presented, one might use sequences.
Without giving away everything, suppose $x_n \rightarrow x$, and consider the the sequence $f_n:=f(x_n)$. If any subsequence $f_{n_k}$ of $f_n$ converges, the only candidate for it to converge to is $f(x)$ by the closedness of the graph. Considering $f_n$ is bounded, what can be said about $\limsup f_n$ and $\liminf f_n$?