Extensions of Functions with a Closed Graph and Quasi-continuous Functions with a Closed Graph from Dense Subspaces

Abstract

In this paper, we prove some conditions for existence of an extension of a real (quasi-continuous) function with a closed graph defined on a given dense subset D of a topological space X to a (quasi-continuous) function with a closed graph on whole X.

1 Introduction

We use mostly standard terminology and notation. Let X be a topological space. For each set $A\subset X$, the symbols ${{\,\textrm{int}\,}}A$ and ${{\,\textrm{cl}\,}}A$ denote the interior and the closure of A, respectively. The spaces ${\mathbb R}$ and $X\times {\mathbb R}$ are considered with their standard topologies. In this paper every considered function is real. If $f: X \rightarrow {\mathbb R},$ the symbol G(f) denotes the graph of f, i.e., $G(f) = \{ (x, f(x)) \,:x \in X \}$. We say that a function $f:X \rightarrow {\mathbb R}$ has a closed graph, if the graph of f is a closed subset of the product $X\times {\mathbb R}$. We say that a function $f:X \rightarrow {\mathbb R}$ is lower (upper) semicontinuous at a point $x\in X$, if for each $\varepsilon >0,$ there is a neighbourhood U of x such that $f(z)>f(x) - \varepsilon$ ($f(z)<f(x)+\varepsilon$, respectively) for each $z\in U$. If $f:X \rightarrow {\mathbb R}$ is lower (upper) semicontinuous at each point $x\in X$, we say that f is lower (upper) semicontinuous. We say that a function $f:X \rightarrow {\mathbb R}$ is quasi-continuous at a point $x\in X$, if for each $\varepsilon >0$ and for each neighbourhood U of x,  there is a nonempty open set $V \subset U$ such that $|f(x) - f(z)| < \varepsilon$ for each $z \in V$. If $f:X \rightarrow {\mathbb R}$ is quasi-continuous at each point $x\in X$, then we say that f is quasi-continuous (see e.g. [10]). The symbols ${lsc}(X)$, ${usc}(X)$, ${{\mathcal {Q}}}(X)$, ${{\mathcal {C}}}(X)$ and ${{\mathcal {U}}}(X)$ denote the class of all real-valued functions defined on X that are lower semicontinuous, upper semicontinuous, quasi-continuous, continuous and have a closed graph, respectively.

The problem of extensions of real functions, inspired by the Tietze’s theorem, was studied in many papers, among others in [5,6,7,8] and [12]. In particular, in 2015,  Wójtowicz and Sieg proved (see [12, Theorem 3.2]) that if X is a topological Hausodorff space and A is a nonempty zero-subset of X (i.e. $A= g^{-1} (0)$ for some continuous function $g:X \rightarrow {\mathbb R}$), then every function $f_0\in {{\mathcal {U}}}(A)$ has an extension $f\in {{\mathcal {U}}}(X)$. Moreover, in [5], some conditions under which a function defined on a topological space X can be semicontinuous were examined. This conditions refer to the behaviour of the function both on and off a dense or closed subset of X. In [7], theorems concerning the extension of semicontinuous and quasi-continuous functions were proven, in particular conditions for the existence of a lower (upper) semicontinuous (quasi-continuous) extension of a function defined on a dense subset of a Hausdorff topological space X to a lower (upper) semicontinuous (quasi-continuous) function on X were given ( [7, Theorems 2.1, 2.2, 2.4 and 2.5]).

In this paper, we study similar problems, but for (quasi-continuous) functions with a closed graph. We prove a sufficient condition for the existence of an extension of a function with a closed graph on a given dense subset of a metric space X to a function with a closed graph on X (see Theorem 2.2). Moreover, we give equivalent conditions for the existence of a (unique) extension of a quasi-continuous function with a closed graph defined on a given dense subset of a Hausdorff topological space X to a quasi-continuous function with a closed graph on X (see Theorem 2.4).

The next part of this paper was inspired by the results of [5, Proposition 3.2] for lower semicontinuous functions. In this paper, we consider similar problems for (quasi-continuous) functions with a closed graph. It is easy to see that if D is a dense subset of a topological space X, $f:X \rightarrow {\mathbb R}$ and the restrictions $f {\mathord {\restriction }}D \cup \{x\}$ are continuous for $x \in X {\setminus } D$, then f is continuous (see also [3]). The analogous theorems are true for quasi-continuous functions (see Lemma 2.6) and quasi-continuous functions with a closed graph (see Proposition 2.7). Ferrer, Gregory and Alegre showed that a similar result holds for semicontinuous functions with some strong assumptions about the sets D and $X{\setminus } D$ (see [5, Proposition 3.2 and Examples 3.3 (a) and (b)]). In this paper, we obtain an analogous result for a function with a closed graph (see Theorem 2.5 and Examples 1 and 2).

2 Main Results

We start our main section with a useful criterion for checking if the given mapping has a closed graph (e.g. [9, Theorem 14.1.1] or [11] p. 191).

Proposition 2.1

Let X and Y be topological spaces. A function $f: X \rightarrow Y$ has a closed graph if and only if for each $x\in X$ and for each net $(x_\gamma )\subset X$, if $x_\gamma \rightarrow x$ and $f(x_\gamma ) \rightarrow y,$ then $y=f(x)$.

We will prove now a sufficient condition for extending functions with a closed graph from a dense subset of a metric space.

Theorem 2.2

Let (X, d) be a metric space and let D be dense in X. Let also $f_0\in {{\mathcal {U}}}(D)$ be a function with a closed graph such that the following conditions hold true:

• (1) for each $x \in X {\setminus } D$ and each sequence $(x_n), (z_n)\subset D$ such that $x_n \rightarrow x$ and $z_n \rightarrow x$, if $f_0(x_n) \rightarrow y\in {\mathbb R}$ and $f_0(z_n) \rightarrow y'\in {\mathbb R}$, then is $y= y'$,

• (2) the set

  $$\begin{aligned} A{\mathop {=}\limits ^{df }}\{x \in X{\setminus } D:\text { if }x_n \rightarrow x\text { then }|f_0(x_n)| \rightarrow \infty \text { for each sequence }(x_n) \subset D\} \end{aligned}$$

  is an $F_\sigma$-subset of X.

Then there exists a function with a closed graph $f:X \rightarrow {\mathbb R}$ such that $f(x)=f_0(x)$ for every $x\in D$.

Proof

For each $x \in X {\setminus } (D \cup A)$ choose $y_x \in {\mathbb R}$ such that there exists a sequence $(x_n)$ of elements of D with $x_n \rightarrow x$ and $f_0(x_n) \rightarrow y_x$.

Now, choose an increasing sequence $(F_n)$ of closed subsets of X such that $A=\bigcup _{n \in {\mathbb N}} F_n$ (recall that A is an $F_\sigma$-subset of X). Put $F_0= \emptyset$ and $E_n = F_n {\setminus } F_{n-1}$ for each $n \in {\mathbb N}$. Define the function $f:X \rightarrow {\mathbb R}$ by the formula:

$$\begin{aligned} f(x)={\left\{ \begin{array}{ll} f_0(x),&{} \text {if }\,x \in D,\\ y_x,&{} \text {if }\,x\in X {\setminus } (D \cup A),\\ 0,&{} \text {if}\, x \in E_1,\\ 1/d(x,F_{n-1}),&{} \hbox {if}\, x \in E_n {,} n \in {\mathbb N}{\setminus } \{1\}. \end{array}\right. } \end{aligned}$$

Clearly, f is an extension of $f_0$. We will show that $f \in {{\mathcal {U}}}(X)$. Fix $x_0\in X$. Let $(x_n)$ be a sequence of elements of X such that $x_n \rightarrow x_0$ and $f(x_n) \rightarrow y\in {\mathbb R}$. We will show that $y=f(x_0)$. Let us consider the following two cases.

Case 1.  $x_0\in A$.

Then $x_0 \in E_k$ for some $k \in {\mathbb N}$. First, observe that for each sequence $(x_n)$ of elements of $X{\setminus } A$, if $x_n \rightarrow x_0,$ then $|f(x_n)| \rightarrow \infty$. Indeed, suppose that there is a sequence $(x_n) \subset X {\setminus } A =D\cup \bigl (X {\setminus } (D \cup A)\bigr )$ such that $x_n \rightarrow x_0$ and $f(x_n) \rightarrow y$ for some $y\in {\mathbb R}$. Then, for each $n \in {\mathbb N}$ such that $x_n \in X {\setminus } (A \cup D)$, there is a sequence $(x^{n}_{m})\subset D$ such that $x^{n}_{m} \rightarrow x_n$ and $f(x^{n}_{m})=f_0(x^{n}_{m}) \rightarrow y_{x_n} = f(x_n)$. Thus, there is a sequence $(x_k)\subset D$ such that $x_k \rightarrow x_0$ and $f(x_k)=f_0(x_k)\rightarrow y$, a contradiction (recall that $x_0\in A$). So, since $x_n \rightarrow x_0$ and $f(x_n) \rightarrow y$, we may assume that $x_n \in A$ for each $n\in {\mathbb N}$ (the proof of the next part of Case 1 is similar to the part of the proof of [4, Theorem 1]). Now, notice that there is $n_0$ such that $x_n \in F_k$ for each $n>n_0$. Indeed, if $x \in \{x_n: n \in {\mathbb N}\}{\setminus } F_k$, then $x \in E_i$ for some $i \ge k+1$ and

$$\begin{aligned} f(x)=1/d(x,F_{i-1})\ge 1/ d(x,x_0). \end{aligned}$$

Since $x_n \rightarrow x_0$ and $f(x_n) \rightarrow y$, there is $n_0$ such that $x_n \in F_k$ for each $n>n_0$.

In the case $k=1$, we have $x_0, x_n \in F_1=E_1$ and $f(x_n)=0$ for each $n>n_0$. Consequently, $f(x_n) \rightarrow f(x_0)=0$ and $y=f(x_0)$.

Now, suppose that $k \ge 2$. Since $x_0 \in E_k=F_k {\setminus } F_{k-1}$ and the set $F_{k-1}$ is closed, there is $n_1 \in {\mathbb N}$ such that $x_n \notin F_{k-1}$ for each $n > n_1$. Thus, $x_n \in E_k$ for each $n> n_0+n_1$. Consequently, since $f {\mathord {\restriction }}E_k$ is continuous, we have $f(x_n) \rightarrow f(x_0)$ and $y=f(x_0)$.

Case 2.  $x_0 \in X {\setminus } A$.

First, observe that for each sequence $(x_n)$ of elements of A, if $x_n \rightarrow x_0,$ then $f(x_n) \rightarrow \infty$. Indeed, suppose that there is a sequence $(x_n) \subset A$ such that $x_n \rightarrow x_0$ and $f(x_n) \rightarrow y\in {\mathbb R}$. Observe that if there exists $k \in {\mathbb N}$ and $n_0 \in {\mathbb N}$ such that $x_n \in F_k$ for each $n>n_0$, then (since $x_n \rightarrow x_0$ and $F_k$ is closed) $x_0 \in F_k \subset A$, a contradiction. Thus, there is a subsequence $(x_{n_m})\subset (x_n)$ such that for each $m_1,m_2 \in {\mathbb N}$, if $m_1 \ne m_2$ there is $k \in {\mathbb N}$ such that $x_{n_{m_1}} \in F_k$ and $x_{n_{m_2}} \in E_i$ for some $i \ge k+1$, which yields

$$\begin{aligned} f(x_{n_{m_2}})=1/d(x_{n_{m_2}},F_{i-1})\ge 1/ d(x_{n_{m_2}},x_{n_{m_1}}), \end{aligned}$$

a contradiction (recall that sequences $(x_{n_m})$ and $f(x_{n_m})$ are convergent).

So, we may assume that $x_n \in X {\setminus } A=D\cup \bigl (X {\setminus } (A \cup D)\bigr )$ for each $n\in {\mathbb N}$. Now, observe that for each $n \in {\mathbb N}$ such that $x_n \in X {\setminus } (A \cup D),$ there is a sequence $(x^{n}_{k})$ of elements of the set D such that $x^{n}_{k} \rightarrow x_n$ and $f(x^{n}_{k})=f_0(x^{n}_{k}) \rightarrow y_{x_n} = f(x_n)$. Thus, there is a sequence $(x_m)$ of elements of D such that $x_m \rightarrow x_0$ and $f(x_m)=f_0(x_m)\rightarrow y$ (recall that $x_n \rightarrow x_0$ and $f(x_n) \rightarrow y$, where $y \in {\mathbb R}$ and $x_n \in D\cup \bigl (X {\setminus } (A \cup D)\bigr )$ for each $n\in {\mathbb N}$).

If $x_0\in X{\setminus }(A \cup D),$ then (by the definition of f and the assumption (1)) $y= y_{x_0} = f(x_0)$. Moreover, if $x_0\in D$, we clearly have $y= f_0 (x_0)=f(x_0)$. Thus $f \in {{\mathcal {U}}}(X)$ and the proof is complete. $\square$

In the next part of the paper, we will use the following characterization of functions with a closed graph (see [2, p. 118] and [1]).

Proposition 2.3

Let X be a topological space. A function $f:X \rightarrow {\mathbb R}$ has a closed graph if and only if for each $x\in X$ and for each $m \in {\mathbb N}$ there is a neighbourhood V of x such that $f(V)\subset (-\infty , -m) \cup (f(x)-1/m, f(x)+ 1/m) \cup (m, \infty )$.

Our next result is a characterization of extending a quasi-continuous function with a closed graph defined on a dense subset of a Hausdorff space X.

Theorem 2.4

Let X be a Hausdorff topological space and let D be dense in X. Let also $f_0:D \rightarrow {\mathbb R}$ be a quasi-continuous function with a closed graph. Then the following three conditions are equivalent:

• 1. there is a unique extension of $f_0$ to a quasi-continuous function with a closed graph $f:X \rightarrow {\mathbb R}$,

• 2. there exists an extension of $f_0$ to a quasi-continuous function with a closed graph $g:X \rightarrow {\mathbb R}$,

• 3.

  • (i) for each $x \in X {\setminus } D$ there exists a net $(x_\gamma )\subset D$ such that $x_\gamma \rightarrow x$ and $f_0(x_\gamma ) \rightarrow y\in {\mathbb R}$,

  • (ii) for each $x \in X {\setminus } D$ and for each nets $(x_\gamma ), (z_{\gamma })\subset D$ such that $x_\gamma \rightarrow x$, $z_{\gamma } \rightarrow x$ and $f_0(x_\gamma ) \rightarrow y\in {\mathbb R}$, $f_0(z_{\gamma }) \rightarrow y'\in {\mathbb R}$, then is $y= y'$.

Proof

The implication (1) $\Rightarrow$ (2) is obvious.

(2) $\Rightarrow$ (3). Since $g:X \rightarrow {\mathbb R}$ is a quasi-continuous extension of $f_0$ and the set D is dense in X, the first requirement of the condition (3) is fulfilled (see [7, Theorem 2.3]. Moreover, since the function g has a closed graph and g is a extension of $f_0$, the second requirement of (3) is obvious.

(3) $\Rightarrow$ (1). Assume that the (3) is true. For each $x \in X {\setminus } D$ choose $y_x \in {\mathbb R}$ such that there exists a net $(x_\gamma )\subset D$ with $x_\gamma \rightarrow x$ and $f_0(x_\gamma ) \rightarrow y_x$.

Let $f:X \rightarrow {\mathbb R}$ be a mapping given by the formula:

$$\begin{aligned} f(x)={\left\{ \begin{array}{ll} f_0(x),&{} \text {if}\, x \in D,\\ y_x,&{} \text {if}\, x\in X {\setminus } D. \end{array}\right. } \end{aligned}$$

Clearly, f is an extension of $f_0$. Moreover, from the proof of [7, Theorem 2.3], it follows that $f \in {{\mathcal {Q}}}(X)$. We must show that $f \in {{\mathcal {U}}}(X)$. Let $x_0\in X$ and $m \in {\mathbb N}$. We will prove that there is a neighbourhood U of $x_0$ such that

$$\begin{aligned} {f(U)\subset (-\infty , -m) \cup (f(x_0)-1/m, f(x_0)+ 1/m) \cup (m, \infty ).} \end{aligned}$$

First, we will show that there is a neighbourhood U of $x_0$ such that $f(x) \in (-\infty , -1/(2m))\cup (f(x_0) - 1/(2m), f(x_0) + 1/({2\,m}) \cup (2\,m, \infty )$ for each $x \in U \cap D$. We consider two cases.

If $x_0 \in D$, then since the function $f {\mathord {\restriction }}D$ has a closed graph the above condition holds.

Now, let $x_0 \in X {\setminus } D$. From the first part of the condition (3) and the definition of the function f, it follows that there is a net $(x_\gamma )$ of elements of D such that $x_\gamma \rightarrow x_0$ and $f_0(x_\gamma )= f(x_\gamma )\rightarrow y_{x_0}=f(x_0)$. Suppose, by way of contradiction, that for each neighbourhood U of $x_0$ there is $x_U \in U \cap D$ such that $f(x_U) \in [-2\,m, 2\,m]$ and $f(x_U) \notin (f(x_0)-1/(2m), f(x_0)+ 1/(2m))$. Consequently, there is a net $(x_\alpha )$ of elements of D such that $x_\alpha \rightarrow x_0$ and $f(x_\alpha )=f_0(x_\alpha )\rightarrow y$, where $y \in [-2m, 2m] {\setminus } (f(x_0)-1/(2m), f(x_0)+ 1/(2m))$. Clearly, $y \ne y_{x_0}=f(x_0)$, a contradiction with the second part of the condition (3). Therefore, there is a neighbourhood U of $x_0$ such that $f(x) \in (-\infty , -2\,m)\cup (f(x_0) - 1/(2m), f(x_0) + 1/(2m)) \cup (2\,m, \infty )$ for each $x \in U \cap D$.

Now, suppose that $f(z) \notin (-\infty , -m)\cup (f(x_0) - 1/m, f(x_0) + 1/m) \cup (m, \infty )$ for some $z \in U {\setminus } D$. Since f is quasi-continuous at z and the set D is dense in X, there is a nonempty open set $V\subset U$ and a point $d \in V \cap D \subset U \cap D$ such that $|f(d)-f(z)| < 1/(2m)$. Observe now that $|f(d)| \le |f(d)-f(z)|+|f(z)|< 1/({2m})+m< 2m$ and $|f(d)-f(x_0)|\ge |f(z)-f(x_0)|- |f(z)-f(d)|> 1/m - 1/({2m})=1/(2m)$. Consequently, $f(d) \notin (-\infty , -2\,m) \cup (f(x_0) - 1/(2m), f(x_0) + 1/(2m))\cup (2\,m, \infty )$, a contradiction.

The last thing is to show that the function f is unique. Let $g:X \rightarrow {\mathbb R}$ be an extension of $f_0:D \rightarrow {\mathbb R}$ such that $g \ne f$. Then there is a point $x_0 \in X{\setminus } D$ such that $g(x_0) \ne f(x_0) =y_{x_0}$. Since $(x_0, f(x_0)) \in {{\,\textrm{cl}\,}}G(g){\setminus } G(g)$, $g \notin {{\mathcal {U}}}(X)$. Thus, f is a unique quasi-continuous extension with a closed graph of $f_0$. That completes the proof. $\square$

The last part of the paper was inspired by the results of [5, Proposition 3.2] for lower semicontinuous functions. We prove an analogous result for functions with a closed graph (the proof of Theorem 2.5 repeats the proof of [5, Proposition 3.2]).

Theorem 2.5

Let X be a topological space and let $f:X \rightarrow {\mathbb R}$. Assume that D is a dense subset of X such that $X {\setminus } D$ is discrete in X, and for each $d \in D$ there is a neighbourhood of d whose intersection with $X {\setminus } D$ is finite. Then f has a closed graph if and only if the restriction $f {\mathord {\restriction }}D \cup \{x\}$ is a function with a closed graph for each $x \in X {\setminus } D$.

Proof

Suppose that $f {\mathord {\restriction }}D \cup \{x\}$ is a function with a closed graph for each $x \in X {\setminus } D$. Let $x_0\in X$ and $m \in {\mathbb N}$. We will show that there is a neighbourhood V of $x_0$ such that

$$\begin{aligned} {f(V)\subset (-\infty , -m) \cup (f(x_0)-1/m, f(x_0)+ 1/m) \cup (m, \infty ).} \end{aligned}$$

(2.1)

We consider two cases. Assume first that $x_0 \in X {\setminus } D$. Since $x_0$ is an isolated point in $X {\setminus } D$, there is an open set U such that $U \cap (X {\setminus } D) = \{x_0\}$. Moreover, since $f {\mathord {\restriction }}D \cup \{x_0\}$ is a function with a closed graph, there is a neighbourhood $V\subset U$ of $x_0$, such that condition (2.1) holds.

On the other hand, let $x_0 \in D$. If $x_0 \in {{\,\textrm{int}\,}}D$, then (since $f {\mathord {\restriction }}D$ has a a closed graph) there is a neighbourhood V of $x_0$ such that condition (1) holds. If $x_0 \in D{\setminus }{{\,\textrm{int}\,}}D$, there is a neighbourhood U of $x_0$ such that $U \cap (X {\setminus } D)=\{c_1,\ldots , c_n\}$. Since $f {\mathord {\restriction }}D \cup \{c_i\}$ is a function with a closed graph for each $i \in \{1,\ldots ,n \}$, there is a neighbourhood $V_i$ of $x_0$ such that $f(z)\in (-\infty , -m) \cup (f(x_0)-1/m, f(x_0)+ 1/m) \cup (m, \infty )$ for each $z \in V_i\cap (D \cup \{c_i\})$ and each $i \in \{1,\ldots ,n \}$. Then $V {\mathop {=}\limits ^{df }}U \cap V_1 \cap \ldots \cap V_n$ is a neighbourhood of $x_0$ which satisfies the condition (2.1). $\square$

The following two examples show that the assumptions about sets D and $X{\setminus } D$ in Theorem 2.5 are necessary

Example 1

Let $({\mathbb R},d)$ be a metric space, where d denotes the natural metric and let $f:{\mathbb R}\rightarrow {\mathbb R}$ be a function given by the formula

$$\begin{aligned} f(x)={\left\{ \begin{array}{ll} 1,&{} \text {if}\, x\le 0,\\ 1/\left( d(x,\left\{ \frac{1}{n+1}, \frac{1}{n}\right\} )\right) ,&{} \hbox {if}\, x \in \left( \frac{1}{n+1}, \frac{1}{n}\right) , \; n \in {\mathbb N},\\ 0,&{} \text {if }\,x \in \left\{ \frac{1}{n}: n \in {\mathbb N}\right\} \cup (1, \infty ), \end{array}\right. } \end{aligned}$$

and let $D {\mathop {=}\limits ^{df }}{\mathbb R}{\setminus } \{1/n \,:n \in {\mathbb N}\}$.

Clearly, the set D is dense in ${\mathbb R}$. It is easy to see that the function $f {\mathord {\restriction }}D \cup \{1/n\}$ has a closed graph for each $n \in {\mathbb N}$ and the set ${\mathbb R}{\setminus } D= \{1/n \,:n \in {\mathbb N}\}$ is discrete in ${\mathbb R}$. However, each neighbourhood of $0 \in D$ has infinitely many points of ${\mathbb R}{\setminus } D$ and $f\notin {{\mathcal {U}}}(X)$ (since $(0,0) \in {{\,\textrm{cl}\,}}G(f) {\setminus } G(f)$).

Example 2

Consider the function $f: {\mathbb R}\rightarrow {\mathbb R}$ defined in Example 1. Let

$$\begin{aligned} D {\mathop {=}\limits ^{df }}{\mathbb R}{\setminus } (\{1/n \,:n \in {\mathbb N}\}\cup \{0\}). \end{aligned}$$

It is easy to see that the restriction $f {\mathord {\restriction }}D \cup \{x\}$ is a function with a closed graph for each $x \in {\mathbb R}{\setminus } D$. However, for each $d \in D,$ there is a neighbourhood V of d such that $V \cap ({\mathbb R}{\setminus } D) = \emptyset$ (because D is open), but 0 is not an isolated point in ${\mathbb R}{\setminus } D$ and $f\notin {{\mathcal {U}}}(X)$.

To complete the paper, we will show that if in the above Theorem 2.5 we replace the assumptions about the sets D and $X {\setminus } D$ by quasi-continuity of f on $X{\setminus } D$, then again $f:X \rightarrow {\mathbb R}$ has a closed graph (and is also quasi-continuous). First, we will prove the following lemma for quasi-continuous functions, which we will use in the proof of the next proposition.

Lemma 2.6

Let X be a topological space and let D be dense in X. The mapping $f:X\rightarrow {\mathbb R}$ is quasi-continuous if and only if $f {\mathord {\restriction }}D \cup \{x\}$ is a quasi-continuous function for each $x \in X {\setminus } D$.

Proof

Let D be dense in X. Assume that $f {\mathord {\restriction }}D \cup \{x\}$ is a quasi-continuous function for each $x \in X {\setminus } D$. We will prove that $f \in {{\mathcal {Q}}}(X)$.

Fix $x_0 \in X$ and $\varepsilon > 0$. Let U be a neighbourhood of $x_0$. Since the restriction $f {\mathord {\restriction }}D \cup \{x_0\}$ is quasi-continuous, there is a nonempty open set $V \subset U$ such that $|f(x) - f(x_0)| < \varepsilon /2$ for each $x \in V \cap (D \cup \{x_0\})$. We will show that $|f(x) - f(x_0)| < \varepsilon$ for each $x \in V {\setminus } (D \cup \{x_0\})$.

Suppose that $|f(z) - f(x_0)| \ge \varepsilon$ for some $z \in V {\setminus } (D \cup \{x_0\})$. Since the restriction $f {\mathord {\restriction }}D \cup \{z\}$ is quasi-continuous, there is a nonempty open set $G \subset V$ such that $|f(x) - f(z)| < \varepsilon /2$ for each $x \in G \cap (D \cup \{z\})$. Thus, since D is dense in X, there exists a point $d \in G \cap D \subset V \cap D$ such that $|f(d) - f(z)| < \varepsilon /2$. Hence $|f(d) - f(x_0)| \ge |f(z) - f(x_0)| - |f(z) - f(d)| \ge \varepsilon /2$, a contradiction.

Thus $|f(x) - f(x_0)| < \varepsilon$ for each $x \in V$ and $f \in {{\mathcal {Q}}}(X)$. $\square$

Proposition 2.7

Let X be a topological space and let D be a dense subset of X. Then $f:X \rightarrow {\mathbb R}$ is a quasi-continuous function with a closed graph if and only if $f {\mathord {\restriction }}D \cup \{x\}$ is a quasi-continuous function with a closed graph for each $x \in X {\setminus } D$.

Proof

Let D be dense in X. Assume that $f {\mathord {\restriction }}D \cup \{x\}$ is a quasi-continuous function with a closed graph for each $x \in X {\setminus } D$. From Lemma 2.6, we obtain that $f \in {{\mathcal {Q}}}(X)$. It is enough to prove that $f \in {{\mathcal {U}}}(X)$.

Fix $x_0 \in X$ and let $m \in {\mathbb N}$. We will show that there is a neighbourhood U of $x_0$ such that $f(U)\subset (-\infty , -m) \cup (f(x_0)-1/m, f(x_0)+ 1/m) \cup (m, \infty )$.

Since the restriction $f {\mathord {\restriction }}D \cup \{x_0\}$ has a closed graph, there is a neighbourhood U of $x_0$ such that $f(x) \in (f(x_0) - 1/(2m), f(x_0) + 1/(2m)) \cup (-\infty , -2\,m)\cup (2\,m, \infty )$ for each $x \in U \cap (D \cup \{x_0\})$. Suppose that $f(z) \notin (f(x_0) - 1/m, f(x_0) + 1/m) \cup (-\infty , -m)\cup (m, \infty )$ for some $z \in U {\setminus } (D \cup \{x_0\})$. Since $f {\mathord {\restriction }}D \cup \{z\}$ is quasi-continuous at z and the set D is dense in X, there is a nonempty open set $V\subset U$ and a point $d \in V \cap D \subset U \cap D$, such that $|f(d)-f(z)| < 1/(2m)$. Observe that $|f(d)| \le |f(d)-f(z)|+|f(z)|< 1/(2m)+m< 2m$ and $|f(d)-f(x_0)|\ge |f(z)-f(x_0)|- |f(z)-f(d)|> 1/m - 1/(2m)=1/(2m)$. Consequently $f(d) \notin (f(x_0) - 1/(2m), f(x_0) + 1/(2m))\cup (-\infty , -2\,m)\cup (2\,m, \infty )$, a contradiction. That completes the proof. $\square$

An open problem related to the above considerations reads as follows.

Problem 2.8

Give an equivalent condition for existence of an extension of a function $f_0$ with a closed graph defined on a given dense subset of a metric or a topological space X, to a function $f:X\rightarrow {\mathbb R}$ with a closed graph.