1

Let $(X, T_X)$ and $(Y, T_Y )$ be topological spaces and let $f : X → Y$ be a function. Let $x ∈ X$.

I want to show that $f$ is continuous at $x$ if: $f(x) ∈ V$ and $V ∈ T_Y ⇒ x \notin \overline {(f ^{−1}(V ))^c}$ .

Then using this result (or otherwise), to show that $f$ is continuous (on $X$) if and only if $f(\overline{A}) \subseteq \overline{f(A)}$ for any subset $A$ of $X$.

Hmm so does $x \notin \overline {(f ^{−1}(V ))^c}$ imply $x \in \overline{X \setminus f^{-1}[V]}$ which means $V$ is a neighbourhood of $f(x)$ such that $f^{-1}[V]$ is a neighbourhood of $x$? And for the second part, referring to $f$ is continuous at $a$ iff for each subset $A$ of $X$ with $a\in \bar A$, $f(a)\in \overline{ f(A)}$., can I say assuming $f$ is continuous. Given a subset $A$, we have $f(A)\subseteq \overline{f(A)}$, which means $A\subseteq f^{-1}f(A)\subseteq f^{-1}\overline{f(A)}$. This last set is closed, so $\bar A\subseteq f^{-1}\overline{f(A)}$, which means $f(\bar A)\subseteq \overline{f(A)}$.

Then conversely, assuming that $f(\bar A)\subseteq \overline{f(A)}$ for each $A\subseteq X$ , let $F$ be closed in $Y$. Then $f(\overline{f^{-1}(F)})\subseteq \overline{f({f^{-1}(F)})}\subseteq \bar F=F $. Thus $\overline{f^{-1}(F)}\subseteq f^{-1}(F)$. Since the converse always holds, we have $f^{-1}(F)=\overline {f^{-1}(F)}$, so this set is closed and $f$ is continuous.

Henno Brandsma
  • 242,131
Homaniac
  • 1,215
  • $f^{-1}\overline{f(A)}$ is only closed if $f$ is continuous everywhere! You mix local and global continuity in inappropriate ways. – Henno Brandsma Feb 24 '19 at 09:16

1 Answers1

1

You're making it way more complicated that it need to be and you also mix global continuity with local continuity...

We have the condition

$$\forall V \in \mathcal{T}_Y: f(x) \in V \implies x \notin \overline{f^{-1}[V]^\complement}\tag{*}$$

If $f$ is continuous at $x$ then (*) holds: suppose $V$ is open in $Y$ containing $f(x)$, then we have $U$ open in $X$, containing $x$, such that $f[U] \subseteq V$. So $U \cap f^{-1}[V]^\complement = \emptyset$ and this neighbourhood $U$ witnesses that $x \notin \overline{f^{-1}[V]^\complement}$

If (*) holds, $f$ is continuous at $x$: suppose $V$ is an open neighbourhood of $f(x)$. We know by the condition that $x \notin \overline{f^{-1}[V]^\complement}$, so for some open neighbourhood $U$ of $x$ we have that $U \cap f^{-1}[V]^\complement = \emptyset$, or equivalently $f[U] \subseteq V$, as required. having this $U$ for our $V$ makes $f$ continuous at $x$.

So the condition (*) is seen to be a simple reformulation of the local continuity condition at $x$:

$$\forall V \in \mathcal{T}_Y: f(x) \in V \implies (\exists U \in \mathcal{T}_X: x \in U \land f[U] \subseteq V)\tag{c}$$

once you realise that $x$ is in the closure of a set iff every open neighbourhood of $x$ intersects that set and that $U \cap f^{-1}[V]^\complement = \emptyset$ and $f[U]\subseteq V$ are equivalent statements.

Henno Brandsma
  • 242,131
  • Thank you so much for the clarification :) the last sentence about equivalent statements the closure of a set (iff condition) is a standard result right? – Homaniac Feb 24 '19 at 09:39
  • @Homaniac yes, in many texts it's even the definition. – Henno Brandsma Feb 24 '19 at 09:39
  • May i also clarify how to quickly see or verify that $U \cap f^{-1}[V]^\complement = \emptyset$ and $f[U]\subseteq V$ are equivalent? – Homaniac Mar 06 '19 at 17:43
  • @Homaniac Suppose $U \cap f^{-1}[V]^\complement = \emptyset$. Let $y \in f[U]$ so $y=f(x)$ with $x \in U$. By the assumption, $x \notin f^{-1}[V]^\complement$, so $x \in f^{-1}[V]$, hence $y=f(x) \in V$. As $y$ was arbitrary, $f[U] \subseteq V$. That's one direction. – Henno Brandsma Mar 06 '19 at 17:47
  • @Homaniac OTOH, assume $f[U] \subseteq V$. Assume that $x \in U \cap f^{-1}[V]^\complement$ exists. Then $x \in U$ (so $f(x) \in f[U]$) and $x \in f^{-1}[V]^\complement$ or $x \notin f^{-1}[V]$ or $f(x) \notin V$. This means $f(x)$ contradicts the assumed inclusion. So no such $x$ exists. That's the reverse direction. – Henno Brandsma Mar 06 '19 at 17:50
  • Thank you very much for the clarity, by the way does your main answer also offer another way of proving that $f$ is continuous (on $X$) if and only if $f(\overline{A}) \subseteq \overline{f(A)}$? – Homaniac Mar 06 '19 at 17:53
  • @Homaniac look among my most popular answers for my proof of that fact, which is even easier. – Henno Brandsma Mar 06 '19 at 17:55