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Continuity Definition

A function $f: (X, \tau_1) \to (Y, \tau_2)$ is continuous if for every open set $V \in \tau_2$ in $Y$, its preimage $f^{-1}(V)$ is an open set in $X$ (i.e., $f^{-1}(V) \in \tau_1$).

Given Topologies and Function

Space $X = \{a, b, c\}$ with topology $\tau_1 = \{\emptyset, \{a, b\}, X\}$.
Space $Y = \{p, q, r\}$ with topology $\tau_2 = \{\emptyset, \{p, r\}, Y\}$.
Function $f: X \to Y$ defined as $f(a)=p, f(b)=p, f(c)=p$.

1. Check for Continuity of $f$

We must check the preimage of every open set in $Y$ ($V \in \tau_2$).

For $V = \emptyset$: $f^{-1}(\emptyset) = \emptyset$. Since $\emptyset \in \tau_1$, this condition holds.
For $V = \{p, r\}$: $f^{-1}(\{p, r\}) = \{x \in X \mid f(x) \in \{p, r\}\}$. Since $f(a)=p$, $f(b)=p$, $f(c)=p$, then $f^{-1}(\{p, r\}) = \{a, b, c\} = X$. Since $X \in \tau_1$, this condition holds.
For $V = Y$: $f^{-1}(Y) = X$. Since $X \in \tau_1$, this condition holds.

Since the preimage of every open set in $Y$ is open in $X$, the function $f$ is continuous.

2. Show $f$ is Not an Open Map

A function $f$ is an open map if the image of every open set in $X$ is open in $Y$.

Consider the open sets in $X$ (elements of $\tau_1$):

For $U = \emptyset$: $f(U) = f(\emptyset) = \emptyset$. Since $\emptyset \in \tau_2$, this is open.
For $U = \{a, b\}$: $f(U) = f(\{a, b\}) = \{f(a), f(b)\} = \{p, p\} = \{p\}$. Is $\{p\}$ an open set in $Y$? No, because $\{p\} \notin \tau_2$.

Since there exists an open set in $X$ (namely $\{a, b\}$) whose image under $f$ (which is $\{p\}$) is not open in $Y$, the continuous function $f$ is not an open map.

3. Show $f$ is Not a Closed Map

A function $f$ is a closed map if the image of every closed set in $X$ is closed in $Y$.

First, find the closed sets in $X$ and $Y$.

Closed sets in $X$ (complements of open sets in $\tau_1$):

$X \setminus \emptyset = X$
$X \setminus \{a, b\} = \{c\}$
$X \setminus X = \emptyset$

So, the closed sets in $X$ are $\{\emptyset, \{c\}, X\}$.

Closed sets in $Y$ (complements of open sets in $\tau_2$):

$Y \setminus \emptyset = Y$
$Y \setminus \{p, r\} = \{q\}$
$Y \setminus Y = \emptyset$

So, the closed sets in $Y$ are $\{\emptyset, \{q\}, Y\}$.

Now, consider the image of every closed set in $X$ under $f$:

For $A = \emptyset$: $f(A) = f(\emptyset) = \emptyset$. Since $\emptyset$ is closed in $Y$, this holds.
For $A = \{c\}$: $f(A) = f(\{c\}) = \{f(c)\} = \{p\}$. Is $\{p\}$ a closed set in $Y$? No, because $\{p\} \notin \{\emptyset, \{q\}, Y\}$.

Since there exists a closed set in $X$ (namely $\{c\}$) whose image under $f$ (which is $\{p\}$) is not closed in $Y$, the continuous function $f$ is not a closed map.

Conclusion

We have shown that the function $f: X \to Y$ defined above is continuous. However, $f$ is neither an open map nor a closed map, as demonstrated by specific counterexamples for open and closed sets.

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