Show that $\{y:I\longrightarrow Y \mid y(j)=y(i) \;\; \forall j\in I\}$ is closed. Let $I$ be a non-empty set and $Y$ be a Hausdorff space. Fix $i\in I$ and define $$D:=\{y:I\longrightarrow Y \mid y(j)=y(i) \;\; \forall j\in I\}.$$
Show that $D$ is a closed subset of $Y^I:=\{f:I\longrightarrow Y\}$ when endowed with product topology.
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 A: Suppose that $f\in Y^I$ and $\{f_\alpha\}_{\alpha\in M}$ ($M$ is a directed set) be a net in $D$ with $f_\alpha\to f$ in product topology. 
Let there exists some $j\neq i$ such that $f(i)\neq f(j)$. So there exist two open subsets of $Y$ like $U,V$   such that $U\cap V=\emptyset$ and $f(i)\in U, f(j)\in V$. 
There exists an $\alpha_1,\alpha_2\in M$ such that for all $\alpha\geq\alpha_1, \beta\geq\alpha_2$ $f_\alpha(i)\in U$ and $f_\alpha(j)\in V$ (Because $f_\alpha\to f$ in product topology and consequently $f_\alpha(i)\to f(i)$ in subspace topology and the same is true for all $j\in I$).
Let  $\alpha_0\in M$ with $\alpha_0\geq\alpha_1,\alpha_0\geq\alpha_2$ (we can suppose it because $M$ is directed).
Then for all $\alpha\geq\alpha_0, f_\alpha(i)\in U, f_\alpha(j)\in V$. But we know $f_\alpha(i)=f_\alpha(j)$ and so for any $\alpha\ge\alpha_0, f_\alpha(i)\in U\cap V$ which is contradiction. 
Hence $f(i)=f(j)$ for all $j\in I$ which means $f\in D$.
A: Set $\Delta=\{(y,y)\colon y\in Y\}\subset Y\times Y$. It is an easy exercise that $\Delta$ is closed if and only if $Y$ is hausdorff.
Also set $p_j\colon Y^I\to Y$, $p_j(f)=f(j)$. The maps $p_j$ are continuous.
Now 
$$D=\bigcap_{j\in I}(p_j,p_i)^{-1}[\Delta]$$
is an intersection of closed sets and hence closed.
A: Let $Y_{(j)}$ denote the $j$th factor of the product space $Y^{I}$ for $j\in I$ and let $p_j\colon Y^{I}\to Y_{(j)}$ be the usual projection onto the $j$th factor which is continuous by definition of the product topology.
Consider the complement $C=Y^I\setminus D$. We will show that $C$ is open. Let $y$ be an element in $C$ then there exist $j,j'$ such that $y(j)\neq y(j')$ and so as $Y$ is Hausdorff, there exist $U, U'$ such that $y(j)\in U$, $y(j')\in U'$ and $U\cap U'=\emptyset$.
Note that $p_j^{-1}(U)$ is open and $p_{j'}^{-1}(V)$ is open it follows that $C_y=p_j^{-1}(U)\cap p_{j'}^{-1}(V)$ is open, and because $U\cap V=\emptyset$ it follows that $C_y\cap D=\emptyset$. So, $C_y$ is an open subset of $C$ containing $y$. The union of arbitrarily many open sets is open and so $\bigcup_{y\in Y}C_y$ is open, but we see that $\bigcup_{y\in Y}C_y$ contains $y$ for each $y\in Y$ and each component of the union has empty intersection with $D$. It follows that $\bigcup_{y\in Y}C_y=C$ and so $C$ is an open subset of $Y^{I}$, hence $D$ is closed.
