Yes I have seen the explanation of why this is so in quantum mechanical textbooks. However, let's use the identity operator and do the following:  $$\langle x|x'\rangle =\langle x|I|x'\rangle =\int\langle x|x''\rangle\langle x''|x\rangle dx''=\int\delta(x-x'')*\delta(x''-x)dx'\tag{1}$$  (the integral is from $-\infty$ to $+\infty$. Now, let's look at what we have inside the integral. Clearly, the term inside  $$\delta(x-x'')*\delta(x''-x)\tag{2}$$  is nonzero only if $x''=x'=x$. We can restate this as follows:  $$\delta(x-x'')*\delta(x''-x)=0\text{, if  } x\neq x'\tag{3}$$  and  $$\delta(x'-x'')*\delta(x''-x)=\infty \text{ (for certain $x''$ in the region from $-\infty$ to $+\infty$)}\tag{4}$$  if $x=x'$; this means that  $$\delta(x-x'')*\delta(x''-x)=\delta(x'-x).\tag{5}$$  However, if we put $\delta(x'-x)$ instead of $\delta(x-x'')*\delta(x''-x)$ we get the following:  $$\int\delta(x-x'')*\delta(x''-x)dx''=\int\delta(x'-x)dx''=\delta(x'-x)\int dx''\tag{6}$$  (I remind that the integral is from $-\infty$ to $+\infty$). So I get that  $$\langle x|x'\rangle=\delta(x'-x)\int dx''\tag{7}$$  which is not just $\delta(x'-x)$. I would be grateful if someone could tell me where my assumptions are wrong and why.
I've tried to find any solution or proof for $$\langle x| x' \rangle=\delta(x-x'),$$ but I only came to this post:   So I got the information, that the vector $|x\rangle$ form a dirac-normalized basis for the Hilbert Space.   I know that the dirac-delta distribution is defined like this: $$\delta(x-x') = \begin{cases} 0 &amp;\mbox{if } x\neq x' \\ \infty &amp; \mbox{if } x=x' \end{cases},$$ this means that my x' is a point on my x axis where I have my infinite high peak. And also $$\int_{-\infty}^{\infty}dx\cdot \delta(x-x')=1.$$  But how actually correlate this with the scalar product of vectors x, x' in the Hilbert Space that form a so-called 'diracl-normailzed" basis of it?  Can you give me some tips on this please? Or maybe you actually know a link, where this is explained.
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When i try [field]= 'NULL' it selects the features with value = 0 (see picture) Why are z-values replaced with zero after in_memory use in Python? $f,g$ continuous from $X$ to $Y$. if they are agree on a dense set $A$ of $X$ then they agree on $X$ Problem:      Suppose $f$ and $g$ are two continuous functions such that $f: X \to Y  $ and $g : X \to Y $. $Y$ is a a Hausdorff space. Suppose $f(x) = g(x)  $ for all $x \in A \subseteq X $ where $A$ is dense in $X$, then $f(x)  = g(x) $ for all $x \in  X $.   Attempt at a solution:  Put $h(x) = f - g $. Therefore, $h: X \to Y $ is continuous and $Y$ is Hausdorff by hypothesis. Also we know $h(x) = 0 $ for all $x \in A $ such that $A$ is dense in $X$. I want to show that $h(x)$ vanishes everywhere in $X$. We can show $h(x) = 0 $ for all $x \in X \setminus A $. Suppose $h(x) &gt; 0 $ on $X \setminus A$. Pick points $y_1,y_2 \in Y $. 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